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E]^EMENTS

GE OMETR Y:'

CONTAININl(l(

THE FIRST SIX BOOKS OF EUCLID,

wii-H A. '.•;:'.-
• *

SUPPLEMENT

ON THE .

QUADRATURE OF THE CIRCLE, AND THE
GEOMETRY OF SOUDS:

e
. '• - . ,

TO WHICH ARE ADDED,

ELEMENTS OF PLANE AND SPHERICAL
TRieONOMETRY.

JOHN JLAYFAIR, F.R,S. Lond.> Edin.

rROFBrnOlt 09 VX'ffnAL PHILOSOPHY, FOSMERLT OF HATHBMATICS, IN THX

UNIVER8ITT 07 EDIMBUROH.

FROM THE LAST LONDON EDITION, ENLARGED.

NEW-YORK;

W. £. DEAir, PRinTER, TO FHaj^KFORT-STREET.

COLLINS AND' HANNAY ; qOLLINS AND CO. ; AND

N. AND J. WHITE.

1833.

« • <

PREFACE.

It is a remarkable feet in the- history df science, that
the oldest book of Elementary Geometry is still consi-
dered as the best, and that the writings of Eocud, at the
distance of twcT thousand years, contlpue to form the
^ most approved introduction to the mathematical sciences.
"^ This remarkable distinction the Greek Geometer owes
^ not only to the elegance and correctness of his demonstra-
^ tionsi but to an arrangement most happily contrived for
"^ the purpose of instruction, — advantages which, when they
reach a certain eminence, secure the works of an author
Sigainst the injuries of time more effectually than even
originality of invention. The Elements of Eucub, how-
ever^ in passing through the hands of the ancient editors
dinrii^ the decline of science, had suffered some diminu-
tion of their excellence, and much skill and learning
have been employed by the modem mathematicians to
deliver them from blemishes which certainly did not enter
into their -original composition. Of these mathematicians,
Pr. SinBON, as he may be accounted the last, has also
been the most successful, and has left very little room for
the ii^nuit^ of future editors to be exercised in, either by

VI PREFACE.

amending the text of Euclid, or by improving the transla*
tionsfromit

Such being the merits of Dr. Simson's edition, and
the reception it has met with having been every way
suitable, the work now offered to the pubUc will perhaps
« appear unnecessary. And indeed, if the geometer just
named had written with a view of accommodating the
Elements of Euclid to the present state of the mathema-
tical sciences, it is not likely that any thing new in Ele-
mentary Geometry would have been soon attempted. But
hia design was different ; it was his object to restore the
writings of Euclid to their original perfection, and to give
them to Modem Europe as nearly as possible in the state
wherein they made their first appearance in Ancient
Greece. For this imdertaking, nobody could be better
qualified than Dr. Simson ; who, to an accurate knowledge,
of the learned languages, and an mdefatigable spirit of re-
search, added a profound skill in the ancient Greometry,
and an admiration of it almost enthusiastic. Accordingly,
he not only restored the text of Euc ud wherever it had
been corrupted^ but in some cases removed imperfections
that probably belonged to the original work : though his
extreme partiality for his author never permitted him to
suppose that such honour could fall to the share either of
himself, or of any other of the modems.

But, after all this was accomplished, something still re-
mained to be done, since, notwithstanding the acknowledg-

« f

■ pRErA'ce. vii

ed excellence of Euclid's Elements, it conld not be doubt-
ed that some alterations might be made thkt would ac-
commodate them better to a state of the mathematical
sciences, so much more improved and extended than at
the period when they were written. Acc<wpdingly, the
object of the edition now offered to the public, is not so
much to give to the writings of Euclid the form which they
originally had, as that which may at pi^gsent render them
most usefid.

One of the alterations made with this view, respects the
the Doctrine of Proportion, the method of treating which,
as it is- laid down in the fifth of Euclid, has great advan-
tages accompanied with considerable defects ; of which,
however, it mujst be observed, that the advantages are es-
sential, and the defects only accidental. To explain the
nature of the former requires a more minute examination
than is suited to this place, and must therefore be reserved
fov the Notes ; but, in the mean time, it may be remarked,
that no definitioq, except that of Euclid, has ever been
given, from which the properties of proportionals can be
deduced by reasonings, which, at the same time that they
are perfectly rigorous, are also simple and direct. As to
the defects, the prolixness and obscurity that have so often
been complained of in the fifth Book, they seem to arise
chiefly from the nature of the language employed, which
being no other than that of ordinary discourse, cannot
express, without much tediousness and circumlocution,
the relations of mathematical quantities, when taken in

▼IH FftlFACS;

lir uUsoft generality, and when no. asaistanoe can be
received from diagrams. As it is plain that the concise
lai^age of Algebra is directly calculated to remedy this
inconvenieiice, I have endeavoured to introduce it here, in
a very siinple form however, and without changing the na*
lure of the reasoning, or departing in any thing from the
rigour of geometrical demonstration. By this means, the
•teps of the reasoging which were before far separated, are
brought near to one another, and the force of the wh<de is
90 clearly and directly perceived, that I am persuaded no
more difBcuIty will be found in understanding the proposi-
tions, of the fifth Book than those of any other of the Ele-
OieirtB.

In the second Book, also, some algebraic signs have
been introduced, for the sake of representing more really
the addition and subtraction of the rectangles on which the
demonstrations depend. The use of such symbolical wri*
ting, in translating from an original, where no symbols are
used, cannot, I think, be regarded as an unwarrantable Ii>»
berty : for, if by that means the translation is not made
into English, it is made into that universal language so
much sought after in all the sciences, but destined, it would
seem, to be enjoyed only by the mathematical.

The alterations above mentioned are the most material
that have beea attempted on the .books of Edclid. . ' There

■

are, however, a few others, which,though less considerable,
it is^hoped may in some degree facilitate the study of the

PREPACK Vt

Elements. Such are those made oh die definitions in
the first Book, and particularly on that of a straight line.
A new axiom is also introduced in the room of the ISth,
for the purpose of demonstrating more easily some of the
properties of parallel lines,. In the third Booki the re-
marks concerning the angles made by a stra^ht line, and
the circumference of a circle, are left out, as tending to
perplex one who has advanced no farther than the elements
of the science. 'The 27th, 28th, and 29th of the sixth
are changed for easier and more simple proposkions, which
do not materially differ from them, and which answer
exactly the same purpose. Some propositions also have
been added ; biit for a fuller detail concerning these
changes, I must refer to the Notes, in which several of the
more difficult, or more interesting subjectfs of Elementary
Geometry are treated' at considerable length.

The Supplement now added to the Six Books of
Euclid is arranged differently from wliat it was in the first
edition of these Elemetits.

The First of the three Books, into which it is divided,

treats of 4he rectification atid quadrature of the circle,-^

subjects that are often omitted altogether in woriu of thi«

kind. They are omitted, however, «s I cociceive, without

any good reason, because, to measure, the length of tha

simplest of all the carves which Geometry treats of, and

the space contained within it, are problems that c^ft&inljr

lielong to the elements of the science, especially as ihej

PREFACE.

are not more difficult than other propositions which are
i|8ua]]y admitted into them. When I speak of the recti- '
iication of the circle, or of measuring the length of the
circumference, I must not be supposed to mean, thqf a
straight Une is to be made equal to the circumference exacts
ly — a problem which, as is wdl known. Geometry has ne-
ver been able to resolve. All that is proposed is, to de-
termine two straight lines tKat shall differ very little from
one another, not more, for instance, than the four hundred,
and ninety-seventh part of the diameter of the circle, and
of which the one shall be greater than the circumference of
that circle, and the other less. In the same manner, the
(quadrature of the circle is performed only by approx-
imation, or by finding two rectangles nearly equal to one
another, one of them greater, and another less, than the
space contained within the circle.

Iii the second Book of the Supplement, which treats of
the intersection of Planes, I have departed as little as pos-
^ble from £ncLip's method of considering the same sub-
ject in his eleventh Book. The demonstration of the
^urth propositioiv is from Leoendre's Elements of Geo-
metry ; that of the sixth is ne^r, as far as I know ; as is
also the solution of the problem in the nineteenth proposi-

tion,— a problem which; though in itself extremely. simple,
has been omitted by Euclid, and hardly ever treated of, in
fMi elementary fonn, by any geometer*

PRfiFACII* XI

With, respect to the Geometry of Solids, in the third
Book, I have departed from Euclid altogether, with^a
view of rendering it both shorter and more compre-
hensive. This, however, is not attempted by introduc-
ing a m6de of reasoning less rigorous thiui that of the
Greek Geometer : for this would be to pay too dear
even for the time that might thereby be saved; but
it is done chiefly by laying aside a certain rule, which,
though it be not essential to the accuracy of demonstra-
tion, Euclid has thought it proper, as much as possible, to
observe.

The rule referred to, is one which influences the arrange-
ment of his propositions through the whole of the Ele-
ments, viz. That in the demonstration of a theorem, he
never supposes any thing to be done, as any line to be
drawn, or any figure to be constructed, the manner of doing
which he has not previously explained. Now, the only use
of this rule is to prevent the admission of impossible or
contradictory suppositions, which, no doubt, might lead
into error; audit is a rule well calculated to answer that
end, as it does not alloW the* existence of any thing to
be supposed, unless the ttung itself be actually exhibit-
ed. 3ut it is not always necessary to make use of this
defence; for the existence of many, things is obviously
possible, and very far from implying a 'contradii^tioQ,
where the method of actually exhibiting them may be al-
together unknown. Thus, it is jdain, that on any given
figui^ 1^ c& base, a 3<rfid may be coDdtituted> or omceired

XII FRErAor;

to exist, equal in solid contents to aAy giv^n solid, (because
a solid, whatever be its base, as its height may be indefi-
nit^ly varied, is capable of alf d^ees of magnitude, from
BOthing upwardis), and yet it may in many cases be a pro>-
Mem of extreme difficulty to assign the height of such a
solid, and actually to exhibit it. Now, tbis very supposf-
tion, that on a given base a sofid of a given magnitude
may be constituted, is one of those,, by the introduction
•f wbich, the Geometry of solids is much shortened,
while all the real accuracy of the demonstrations is pre^
served ; and therefore, to follow, as JE^c^clid has done,
the rule that excludes this, and such like hypotheses, is
ta- create artificial difficulties, and to embarrass geome-
trical investigation with more obstacles than the nature
of things has thrown in its way. It is a rule, too, which
cannot always be followed, and from which even Eucud
himself has been forced to depart ia more than one in-
stance.

In the Book, therefore, on the Properties of Solids,
, which X now offer to the public, I have not sought to
subject the d^smonstration^ .to the law just mentioned,
and have never hesitated to Admit the existence of such'
solids, or such lines as are evidently possible, though the
manner of actually describing them may not have been'
explained. In this way I have been enabled to offer
that very refined artifice in geometrical reasoning, to
which we give the name of the Method of Exhaustions,
under a much simpler fona than it appears in tiie 12th

PREFACE. ^m

of Euclid ; and the spirit of the method may, I think,
be best learned when it is* tibus diseiigaged from every
thing not essential. That it m«y be the better under-
stood, and because the demonstrations which require ex-
haustions are, no doubt, the most difficult ^in the Elements^
they are all conducted as nearly as possible in the same
way, in the cases of the different solids, from the pyra-
mid to the sphere. The comparison of this last solid
with the cylinder, concludes the last Book of the Sup-
plement, and is a proposition that may not improperly
be considered as terminating the elementary part of Geo-
.metry.

The Book of- the Data has been annexed to several
editions of Euclid's Elements, and particularly to Dr.
Simson's, but in this it i^ omitted altogether. It is omit-
ted, however, not from any opinion of its being in itself
useless, but because it does not belong to this place, and
is not often read by beginners. It contains the rudiments
of what is properly called the Geometrical Analysis, and
has itself an analytical form ; and for these reasons,.!
would vrillingly reserve it, or vather a compend of it, for
a separate work, intended as an introduction to the study
of that analysis.

In explaining the elements of Plane and Spherical
Trigonometry, there is not much new that can be at-
tempted, or that will be expected by the intelligent read-
er. Except, perhaps, some new demonstrations,' and

ZIV PREFACS.

some changes in the arrangement, these two treatises
have, accor%ngIy» no novelty, fo boast of. The Plane Tri-
gonometry is so divided, that Jhe part of it that is barely
sufficient for the resolution of Triangles may be easily
taught by itself.. The method of constructing the Trigono-
metrical Tables is explained, and a demonstration is add-
ed of those properties of the sines and cosines of arches,
tvhich are the foundation of those^ applications of Trigono-
metry lately introduced, with so much advantage, into the
higher Geometry.

In the Spherical Trigonometry, the rules for preventing
the ambiguity of the solutions, wherever it can be pre-
vented, have been particularly attended to ; and I have
availed myself as much as possible of that excellent ab-
stract of the rules of this science which Dr. Maskielyne
has prefixed to Taylor's Tables of Logarithms.

An explanation of Napier's very ingenious and useful
rule of the Circular Parts is here added as an Appendi^i
to Spherical Trigonometry.

It has been objected to many of the writers on Ele-
' mentary Geometry, and particularly to Euclid, that they
have been at great pains to prove the truth of many sim-
pie propositions, which every body is ready to admit
without any demonstration, and that thus they take up the
time, and fatigue the attention of the student, to no pur-
pose* To this objection^ if there be any force in it, the

PREFirCS. Xf

* . ■....* ,

present treatise is certainly as much exposed as any other ;
for no attempt is here made to abridge the Elements,
by considering as self-evident any thing that admits of
being proved. Indeed, those who make the objection just
stated, do not seem to have reflected sufliciently on the
end of Mathematical Demonstration, which is not only to
prove the truth of a certain proposition, but to shew its
necessary connection with other propositions, and its de-
pendence on them. The truths of Geometry are all ne-
cessarily connected with one another, and the system of
such truths cai^ never be rightly explained, unless that con-
nection be accurately traced, wherever it exists. It is upon
this that the beauty and peculiar excellence of the mathe-
matical sciences depend : it is this, which, by preventing
any one truth from, being single and insulated, connects
tjjie different parts so firmly, that they must all stand, or all
fall together. The demonstration, therefore, even of an
obvious proposition, answers the purpose of connecting
that proposition with others, and ascertaining its place in
the general system of mathematical truth. If, for example^
it be alleged, that it is needless to demonstrs^te that any
two sides of a triangle are greater thai^ the third ; it may
be replied, that this is no doubt a truth, which, -without
priDof, most men will be inclined to admit ; but are we for
that reason to account It of no consequence to know what
the propositions are, which must cease to be true if this
proposition WQre supposed to be false ? Is it not useful to
know, that unless it bo true, that any two sides of a tri-

SLVl . PHEFAGB.

imgle are greater than the third, nekher could^ it be true
Ifaat the greater side of every triangle is opposite to the
greater angle, nor that the equal sides are opposite to equal
angles, nor lastly, that things equal to the same thing are
equal to one another ? By a scientific mind this informa-
tion will not be thought lightly of; and it is exactly tfiat
. which we^receive from Euclio^s demonstration.

To all this it may be added, thact the mind, especially
when beginning to study the art of reasoning, cannot be
employed to greater advantage than in analysing those
judgments, which, though they appear simj^le, are in rea-
lity complex, and capable of being distinguished into parts.
No progress in ascending higher can be expected, till a
regular habit of demonstration is thus acquired ; it is much
to be feared, that he who has decKneJ the trouble of trac-
ing the connection between the proposition already quoted*,
and those that are more simple, will not be very expert in
tracing its connection with those that are more complex ;
and that, as he has not been careful in laying the foundation,
he will never be successful in raising the superstructure.

College of Edi^bubgh,
Dec: If 1813,

SlLfiMENTS

GEO ME TRY.

BOOK I.

DEFINITIONS.

** A Point is that which has position, but not magnitude".'* (Sed
Notes.)

^ * , ' IL

A line is length without breadth.

^' CoBOLLARY. The extremities of a. line are points ; and the intet*
** sections of one line w:ith another are also points.'*

III.

** If two lines are such that they cannot coincide in any two points,
" without coinciding altogether, each of them is called a straight
"line."

^ Cos. Hence two straight lines cannot inclose a space. Neither can
' ** two straight lines have a common segment ; that is, they cannot
<< coincide in part, without coinciding altogether*"

IV.

A superficies is that which has only length and breadth.
"^^ CoR« The extremities of a superficies are hnes ; and the intersec*
'^tions of one superficies with another are also lines."

A plane superficies is that in which any two points being taken, the
straight line between^ them lies wholly in that superficies^

VI.

A plane rectilineal angle is the inclination of two straight lines to one
another, which meet together, but are not in the same straight line*

* The defioitioDft marked witli ioverted commas are different from those of £uclid*

ELEMENTS

N. B. ' When several angles are at one point B, any one of them

* is expressed by three letters, of which the letter that is at the ver-'
' tex of the angle, that is, at the point in which the straight lines that
^ contain the angle meet one another, is put between the other two
' letters, and one of these two is somewhere upon one of those straight

* lines, and the other upon the other line : Thus the angle which
' is contained by the straight lines, AB, CB, is named the angle

* ABC, or CBA ; that which is contained by AB, BD, is named the

* angle ABD, or DBA ; and that which is Contained by BD, CB, i»
< called the angle DEC, or CBD ; but, if there be only one angle at

* a point, it may be expressed by a letter placed at that point ; as the

* angle at E.'

VIL

When a straight line standing on another
straight line makes the adjacent angles
equal to one another, each of the angles
is called a right angle ; and the straight
line which stands on the other, is called
a perpendicular to it.

VIII.

An obtuse angle is that which is greater than a right angle*

IX.

An acute angle is that which is less than a right angle.

A figure is that which is inclosed by one or more boundaries. — The
word area denotes the quantity of space contained in a figure, toithouf
any reference to the nature of the line or lines which bound it*

OF GEOMETRY. BOOK I.

A circle is a plane figure contained by one line, which is called the
circucnfcrence, and is such that all straight lines drawn from a cer-
tain point within the figure to the circumference, are equal to one
another. ,^

XII.

And this point is called the centre of the circle.

XIIL
A diameter of a circle is a straight line drawn through the centre, and
terminated both ways by the circumference.

XIV.
A semicircle is the figure contained by a diameter and the part of the
circumference cut off by the diameter.

XV.
Rectilineal figures are those which are contained by straight lines.

, XVI.
Trilateral figures, or triangles, by three straight lines.

XVII.
Quadrilateral, by four straight lines.

XVIII.
Multilateral figures, or polygons, by more than four straight lines.

XIX.
Of three sided figures, an equilateral triangle is that which has three
equal sides.

XX.
An isosceles triangle is that which has only two sides equal.

XXI.

A scalene triangle is that which has three unequal sides.

XXII.
A right angled triangle is that which has a right angle.

ELEMENT}?

XXIII.
An obtuse angled triangle, is that which has an obtuse angfe.

XXIV.
An acute angled triangle, is that which has three acute angles.

XXV.

Of four sided figures, a square is that which has all its sides equal
and all its angles right angles.

XXVL

An oblong, is that which has all its angles right angles, but has no^alt
its sides equal.

XXVIt.

A rhombus, is that which has all its sid^s equal, but its angles are net
right pmgles^

xxvm.

A rhomboid, is that which has its opposite sides equal to one another,,
but all its udes are not equal, nor its angles right angles.

XXIX.

All other four sided figures besides these, are called Trapeziums..

XXX.

Parallel straight lines, are such as are in the same plane, and which,,
being produced ever so far both ways, do not meet..

OF GEOMETRt. BOOK I. 31

POSTULATES.
I.

Let it be granted that a straight line may be drawn from any one
point to any other point.

That a terminated straight line may be produced to any length in a
Straight line.

III.

And that a circle may be described from any centre, at any distance
from that centre.

AXIOMS.
I.

Things which are equal to the same thing are equal to one another*

11.
If equals be added to equals, the wholes are equal.

III.

If equals be taken from equals, the remainders are equal.

IV.

If equals be added to unequals, the wholes are unequal.

If equals be taken from unequals, the remainders are unequal.

vi.

Things which are doubles of the same thing, are equal to one anothen

VIL

Things which are halves of the same thing, are equal to one another.

VIII.
Magnitudes which coincide with one another, that is, which exactly
fill the same space, are equal to one another.

IX.

The whole is greater than its part.

All right angles are equal to one another. ,

XL
" Two straight lines which intersect one another, cannot be both pa-^
*' rallcl to tho same straight line."

2Si

ELEMENTS

PROJ^OSITION I. PBOBLEM.
To describe an equUatercd triangle upon a given finite straight line.

Let AB be the given straight line ; it is required to describe an
equilateral triangle upon it.

From the centre A, at the -

distance AB, describe (3. Pos- ^ ^

tulate) the circle BCD, and
from the centre B, at the dis-
tance BA, describe the circle

ACE ; and from the point C, [D Al'^ ^B IE

in which the circles cut one
another, draw the straight
lines (1. Post,) CA, CB to the
points A, B ; ABC is an equi.
lateral triangle.

Because the point A is the centre of the circle BCD, AC is equal
(11. Definition) to AB ; and because the point B is the centre of the
circle ACE, BC is dqual to AB : But it has been proved that CA is
equal to AB ; therefore CA, CB are each of them equal to AB ; now
things which are equal to the same are equal to one another, (1. Ax-
iom) ; therefore C A is equal to CB ; wherefore CA,, AB, CB are
equal to one another ; and the triangle ABC is therefore equilateral,
« and it is described upon the given straight line AB. Which was^ re«
quired to be done.

PROP. IL PROB.

From a given point to draw a straight line equal to a given straight

line* '

Let A be the given point, and BC the given straight line ; it is re-
quired to draw, from the point A, a straight line equal to BC.

From the point A to B draw, (1. Post.)
the straight line AB ; and upon it describe
(1. 1.) the equilateral triangle DAB, and
produce (2. Post.) the straight lines DA,
BD, to E and F ; from the centre B, at
the distance BC,' describe (8. Post.) the
circle CGH, and from the centre D, at the
distance DG, describe the circle GKL.
AL is equal to BC.

Because the point B is the centre of the
circle CGH, BC is eqiAl j(ll. Def.) to
BG ; and because D is the centre of the
circle GEL, DL is equal to DG, and DA,
DB, parts of them^ are equal ; therefore

ma*

OF GEOMETRY. BOOK I.

the remainder AL ts equal to the remaiader (Q« Ax.) B6 : But it ha^
been she^n that BC is equal to BG ; wherefoire AL and BC are each
of them equal to BG ; and things that are eqncd to the same are equal
to on^ another; therefore the straight line AL i& equal to BO.
Wherefore, from the given point A, a straight line AL has been drawn
equal to the given straight line BC. Which was to be done.

PROP. in. PROB.

From the greater of two given straight lines to cut off a part equal io the

less.

Let AB and C be the two given
straight lines, whereof AB is the great-
er. It is required to cut off from AB,
the greater, a part equal to C, the less.

From the point A draw (2. 1.) the
straight line AD equal to C ; and from
the centre A, and at the distance AD,
describe (3, Post.) the circle DEF;
and because A is the centre of the cir-
cle DEF, AE is equal to AD; but the
straight line C is likewise equal to

AD ; whence AE and C are each of them equal to AD ; wherefore
the straight Ime AE is equal to (1. Ax.) C, and from AB the greater
of two straight lines, a part AE has been cut off equal to C the less.
Which was to be done.

PROP. IV. THEOREM.

Jf two triangles have two sides of the one equal to two sides of the other ,.
each to each ; and have likewise the angles contained by tJiose sides
equal to one anotJier^ their bases, or third sides, sJiaU be equal ; and
the areas of the triangles shall be equal ; and their other angles shaU
be equal fea^h to each, viz. those to which the equal sides are opposite.*

Let ABO, DEF be two triangles which have the two sides AB, AC
equal to the two sides DE, DF, each to each, viz. AB to DE> and
AC to DF ; and let the * Ti

angle BAC be also "^ ^

equal to the angle
EDF: then shall the
base BC be equal to
the base EF ; and the
triangle ABC to the tri-
angle DEF; and the
other angles, to which
the equal sides are op- S

* The three concltsions in this enanciation are more briefljr expressed by Baying, that
the triangles are every way equal.

34 BLEMENTS

pofidte, shall be equal, each to each, viz. the angle ABC to the angle
D£F, and the angle ACB to DFE.

For, if the triangle ABC be applied to the triangle DEF, so that
the point A may, be on D, and the straight line AB upon DE ; the
point B shall coincide with the point E, because AB is equal to DE ;
and AB coinciding with DE, AC shall coincide with DF, because the
angle BAC is equal to the angle EDF ; wherefore also the point
shall coincide with the point F, because AC is equal to DF : But the
point B coincides with the point £ ; wherefore the base BC shall
coincide with the base EF (cor. def. 3.), and shall be equal to it.
Therefore also the whole triangle ABC shall coincide with the whole
triangle DEF, so that the spaces which they contain or their areas
are equal ; and the remaining angles of the one shall coincide with
the remaining angles of thS other, and be equal to them, viz. the
angle ABC to the angle DEF, and the angle ACB to the angle DFE.
Therefore, if two triangles have two sides of the one equiA to two
sides of the other, each to each, and have likewise the angles contain-
ed by those sides equal to one another ; their bases shall be equal,
and their areas shall be equal, and their other angles, to which the
equal sides are opposite, shall be equal, each to each. Which was to
be demonstrated.

PROP. V. THEOR.

The angles at the base of an Isosceles triangle are equal to one another ;
and %f the equal sides he produced^ the angles upon the other side of
the base shall be equcd.

Let ABC be an isosceles triangle, of which the side AB is equal to
AC, and let the straight lines AB, AC be produced to D and E, the
angle ABC shall be equal to the angle ACB, and the angle CBD to
the angle BCE.

In BD take any point F, and from AE the greater cut off AG equal
(3. 1.) to AF, the less, and join FC, GB. Cx B>

Because AF is equal to AG, and^ AC, the two sidfes FA, AC
are equal tothe two GA, AB, each roeach ; and they contain the
angle FAG common to the two trian-

fles, AFC, AGB ; therefore the base
'C is equal (4. 1.) to the base GB,
and the triangle AFC to the triangle
AGB ; and the remaining angles of
the one are equal to (4. 1.) to the re.
maining angles of the other, each to
each, to which th^ equal sides are
opposite, viz. the angle ACF to the
ai^gle ABG, and the angle AFC to the
angle AGB : And because the whole
AF is equal to the whole AG, and the
part AB to the part AC ; the remain, jy
der BF shall be equal (3. Ax.) to the

OF GEOMETRY. BOOK I. 25

tH^ainder CG ; and FC was proved to be equal to GB, therefore the
two sides BF, FC are equal to the twoCG, GB, each to each ; but the
angle BFC is equal to the angle CGB ; wherefore the triangles BFO,
OGB are equal (3. l.),-and their remaining angles are equal, to which
the equal sides are opposite ; therefore the angle FBC is equal to
the angle GOB, and the angle BCF to the angle CBG. Now, since
it has been demonstrated, that the^hole angle ABG is equal to the
whole ACF, and the part CBG to the part BCF, the remaining angle
ABC is therefore equaUto the remaining angle ACB, which are the
angles at the base of the triangle A-BG : And it has also been proved
that the angle FBC is equal to the angle GCB, which are the angles
tipon the other side of the base. Therefore, th6 angles at the base,
&c. Q. E. D.
Co(RoirLASY. Hence every equilateral triangle is also equiangular*

PROP. VI. THEOR.

tfttoo angles of a triangle be equal to one auQther, the sides which suh*
teiidy or are opposite to thentf are also equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle
ACB ; the side AB is also equal to the side AC.

For, if AB be not equal to AC, one of them is greater than the
other : Let AB be the greater, and from it cut (3. 1.) off DB equal
to AC the less, and join DC ; therefore, be-
cause ia the triangles DBC, ACB, DB is
equal to AC, and BC common to both, the
two sides DB, BC are equal to the two AC,
CB, each to each ; but the angle DBC is also
equal to the angle ACB ; therefore the base
DC is equal to the base AB, and the area
of the triangle DBC is equal to that of the
triangle (4. 1.) ACB, the less to the greater ;
which is absurd. Therefore, AB is not un-
equal to AC, that is, it is equal to it. Where*
fore, if two angles, dtc. Q. E. D. -B

CoE. Hence every equiangular triangle is also equilateral.

PROP. VIL THEOR.

Vpon the same hose, and on the same side of it, there cannot he two tri*
angles, that ham their sides which are terminated in one extremity of
the base equal to one another^ and likewise those which are terminated
in the other extremity , equal to one another.

Let there be two triangles ACB, ADB, upon the same base AB, and
upon the same side of it, which have their sides CA, DA-; terminated
in A equal to one another ; then their sides CB, DB, terminated in

ELEMENTS

By cannot be equal to one another.

Join CD, and if possible let CB
be equal to DB ; then, in the case
Ui which the vertex of each of the
triangles is without the other trian-
gle, because AC is equal to AD, the
angle ACD is equal (5. 1.) tcithe
angle ADC : But the angle ACD is
creator than the angle BCD ;* there-
tore the angle ADC is greater also
than BCD ;' much more then is the
angle BDC greatef than the angle
BCD. Again, because CB is equal to DB, the angle BDC is equal
(5. 1.) to the angle BCD ; but it has been demonstrated to be greater
than it ; which is impossible.

But if one of the vertices, as ^

Dy be within the other triangle
ACB ; produce AC, AD to E,
F; therefore, because AC is
equal to AD in the triangle
ACD, the angles ECD, FDC
upon the other side of the base
CD are equal (5. 1.) tooiie ano-
ther, but the angle ECD is great- ^
er than the angle BOD ; where-
fore the angle FDC is likewise greater than BCD ; ^much more then
is the angle BDC greater than the angle BCD. A^gain, because CB
is equal to DB, the angle BDC is equal (5. 1.) to the angle BCD ; but
BDC has been proved to be greater than the same BCD ; which is
impossible. The case in which the vertex of one triangle is upon a
side of the other, needs no demonstration.

Therefore, upon the same base, and on the same side of it, there
cannot be two triangles that have their sides which are terminated in
one extremity of the base equal to one anotherj and likewise those
which are terminated in the other extremity equal to one another.
Q. E. D.

PROP. VIII. THEOR.

If two triangles have two sides of the one equal to two sides of the
other y each to each, and have likewise their bases equal ; the angle which
is contained by the two sides of the one shall be equal to the angle con'
tained by the two sides of the other.

Let ABC, DBF be two triangles having the two sides^ AB, AC,
equal to the two sides D£, DF, each to each, viz. AB to DE, and AC

OP GEOMETRYi BOOK I.

to DF ; and also the base BC equal to the base EF. The angle BAC
is equal to the angle EDF.

For, if the triangle ABC be applied to the triangle DEF, so that
the point B be on E, and the straight line BC upon EF ; the point C
shall also coincide with the point F, because BC is equal to £F :
therefore BC coinciding with EF, BA and AC shall coincide with ED
and DF ; for, if BA and CA do not coincide with ED and FD, but
have a different situation, as EG and FG ; then, upon the same base
EF, and upon the same side of it, there can be two triangles EDF,
EGF, that have their sides which are terminated in one extremity of
the base equal to one another, mid likewise their sides terminated
in the other extremity; but this is impossible (7. 1.) ; therefore, if
the base BC coincides with the base EF, the sides BA, AC cannot
but coincide with the sides ED, DF ; wherefore likewise the angle
BAC coincides with the angle EDF, and is equal (8. Ax.) to it.
Therefore, if two triangles, &c. Q. E. D.

PROP. IX. PROB.

To bisect a given rectilineal angle, that is, to divide it into two equal

angles^

Let BAC be the given rectilineal angle, it is required to bisect it.

Take any point D in AB, and from AC cut (3. 1.) off AE equal to
AD ; join DE, and upon it describe
(1. 1.) an equilateral triangle DEF;
then join AF ; the straight line AF bi-
sects the angle BAC.

Because AD is equal to AE, and AF
is common to the two triangled^DAF,
EAF ; the two sides DA, AF, are equal
to'the two sides E A, AF, each to each ;
but the base DF is also equal to the
base EF ; therefore the angle DAF is
equal (8. 1.) to the angle EAF : where-
fore the given rectilineal angle BAC
is bisected by. the straight line AF.
Which was to be done.

ELEMENTS

PROP. X. PROB.

To bisect a gwenfinUe straight linCy that is^ to ditide U into twa eqwi

farts*

Let AB be the given straight line \ it is requhred to divide it into^
two equal parts.

Describe (I. 1.) upon it an equilateral triangle ABC, and biseet
(9. 1») the angle ACB by the straight line CD. AB is cut into twe>
equal parts in the point D.

Because AC is equal to C6, and CD common to the two tciangle»
ACD, BCD : the two sides AC, CD, are ' p

equal to the two BC,. CD, each to each f ^'

but the angle ACD is also equal to the an-
gle BCD ; therefore the base AD is equal
to the base (4. 1.) DB, and the straight
line AB is divided into two equal parts iik
the point D. Which was to be done..

PROP. XI, PROB.

To draw a straight line at right angles to a given stnjight Une^fromr

a given point in that line.

Let AB be a given straight line, and Ca point given in it ; it is re-
quired to draw a straight line from the point C at right angles to A1&

Take any point D in AC, and (^ 1.) make CE equal to CD, aaii
upon DE describe (1. L) the -qi

equilateral triangle DFE, and "*^

j,oin FC ; the straight line FC,
drawn from the given point C,
is at right angles to the given
straight line AB.

Because DC is equal toCE,

and FC common to the two tri- ^^ ^ -c* x»

anglesDCF,ECF, the two sides A- 1> ^ H- JS

DC, CF are equal to the two EC, CF, each to each ; but the base^
DF is also equal to the base EF ; therefore the angle DCF is equal
(8. 1.) to the angle ECF ; and they are adjacent angles. But;^ when
the adjacent angles which one straight line makes with another straight
line are equal to one another, each of them is called a right (7. def.)r
angle ^ therefore each of the angles DCF, ECF, is a[ right angle.
Wherefore, from the givei^ point C> in the given straight line AB>
FC has been drawn at right angles to AB. Which was to be iooj^

OF GEOMETRY. BOOK I.

PROP. Xn. PROB-

To draw a straight line perpendietdar to a given stratght ZthCy of <m
urdimited length, from a given point toithout it* ,

Let AB fee a given straight line, which may he produced to any
length hoth ways, and let G be a point without it. It is required to
draw a straight line perpendi-
cular to AB from the point C.

Take any point D upon the
other side of AB, and from the
centre C, at the distance CD,
describe (3. Post.) the cilrcle
EGF meeting AB in F, 6: and
bisect (10. 1.) FG in H, and
join CF, CH, CG ; the straight
line CH, drawn from the given point C, is perpendicular to the given
straight line AB.

Because FH is equal to HG, and HC common to the two triandes
FHC, GHC, theiwo sides FH, HC are equal to the two GH, HC,
each to each ; but the base CF is also equal (11. Def. I.) to the base
CG ; therefore the angle CHF is equal (8. 1.) to the angle CHG ;
and they are adjacent angles ; now when a straight line standing on
a straight line makes the adjacent angles equal to one another, each
of them is a right angle, and the straight line which stands upon the
other is called a perpendicular to it ; therefore from the given point
C a perpendicular CH has been drawn to the given straight line AJB.
Which was to be done.

PROP. XIII. THEOR.

77^ angles which one straight line makes with another upon one side of
itf are either two right angles, or are together equal to two right an^
gies.

Let the straight line AB make with CD, upon one side of it the
angles CBA, ABD ; these are either two right angles, or are together
equal to two right angles.

For, if the angle CBA be equal to ABD, each of them is a right
angle (Def. 7.) ; but, if not, from the point B draw BE at right an-

A E A

'T^ft

90 ELEMENTS

gles (11. L) to CD ; therefore the angles CBE, EBD are two righf
angles. Now, the angle CBE is equ^ to the two angles CBA, ABE
together ; add the angle EBD to each of these equals, and the two
angles CBE^ EBD will be equal (2. Ax.) to tlie three CBA, ABE,
EBD. Again, the angle DBA is equal to the two angles DBE, EBA ;
add tio each of these equals the angle ABC ; then w3l the two angles
DBA, ABC be equal to the three angles DBE, EBA, ABC ; but the
angles CBE, EBD have been demonstrated to be equal to the same
three angles ; and ihings.that are equal to the same are equal (1 . Ax.)'
to one another ; therefore the angles CBE, EBD are equal to the
angles DBA, ABC ; but CBE, EBD^ are two right ttngles; therefore
DBA, ABC ; a;'e together equal to two right angles. Wherefore,
when a straight line, dsc. Q. £* D«

PROP.XIV! THEOR.

IJT, at a point itp a straigM line, two other straight lineSy upon the oppo^
site sides of it, make the adjacent angles together equal to two right
angles, these Hso straight lines are in one and the same straight line*

At the point B in the straight
line AB, let the two straij^ht lines
Be, BD upon the opposite sides
of AB, make the adjacent an-
gles ABC, ABD equal together
to two right angles. BD is in the
same straight hne with CB.

For if BD be not in the same
straight line with CB, let BE be

in the same straight line with it ; ^ ■ -^j ■ — =l

therefore, because the straight ^ -*^ -D

line AB mi^s angles with the straight line CBE, upon one side of
it, the angles ABC, ABE are together equal (13. 1.) to two right an-
gles ; but the angles ABC, ABD are likewise together equal to two
-right angles: therefore the angles CBA, ABE are equal to the an-
gles CBA, ABD : Take away the common angle ABC, and the re-
maining angle ABE is equal (3. Ax.) to the remaining angle ABD,.
the less to the greater, which is impossible ; therefore BE is not in
the same straight line with BC. And in like manner, it may b^ de«
monstrated, that no other t;an be in the same straight line with it but
BD, which therefore is in the same straight line with CB. Wherefore,,
if at a point, &c. Q. E. D.

PROP. XV. THEOR.

If two straight lines cut one another , the vertical, or opposite angles are

equal.

Let the two straight lines AB, CD, cut one another in the point E t
the angle A£C shall b^ equal to th^ angle DEB, and CEB to AED^

OPGEGMETllT- BOOK I.

»t

lFV>r the angles €EA, A£D,i¥liioh tlie sMight line 'A£ mikkM nvMl
the straight line CD, are togelberequaL(l3.1.) to 4wo right aiiglea:
and the angles AED, DEB, p
which the straight line DE ^
makes with the straight line
AB, are also together equal
(13. 1.) to two right angles ;
therefore the two angles
CEA, AED are equal to the
two AED, DEB. Take a-
' way the common angle AED,
and the remaining angle CEA
is equal (3. Ax.) to the remaining angle DEB. In the same 4naa-
ner it may be demonstrated that the angles CEE, AED are equal*
Therefore, if two straight lines, &c. Q. E. D.

Cor. 1. From this it is manifest, that if two straight lilies cut :on«
another, the angles which they make at the point of their intersection^
' are together equal to four right angles.

Cos. 2. And hence, all the angles made by any number of straighl
lines meeting in one point, are together equal to four light angles.

PROP. XVI. THEQR.

If me side of d triangle, he produced^ the eaieriar angle is greater thm

either of the inferior, and opposite angles.

Let ABC be a triangle, and let its side BC be produced to D, the
exterior angle ACD is greater than either of the interior oj^posite an^
gles CBA, BAC.

Bisect (10. I.) AC in E, join
BE and produce it to F, and
make EF equal to BE ; join al-
2^0 FC, and produce AC to G.

Because AE is equal to EC,
and BE to EF; AE, EB are
equal to C£, EF, each to each ;
and the angle AEB is equal (15.
1.) to the angle CEF, because
they are vertical angles ; there-
fore the base AB is equal (4. 1,)
to the base CF, and the triangle-
AEB to the triangle CEF, and
the remaining angles to the re-
maining angles each to each, to
which the equal sides are oppo«
site; wherefore the angle BAE is equal to the angle EOF ; but the
angle ECD is greater than the angle ECF ; therefore the angle ECl3,
that is ACD^ is greater than BAE : In the same manner, if the side

ELEMENTS

BO be bisected, it may be demonstrated that the angle BCO, th&t i^
(15. L), the angle ACD, is greater than the angle ABC. Therefor^
if one side, &c. Q. E. D.

PROP- XVit THEOR.

Jinp two angles of a triangle are together less than iwo riglU angles^

Let ABC be any triangle ; any
two of its angles together are less
Chan two right angles.

Produce BC to D ; and because
A CD is the exterior angle of the
triangle ABC> ACD is greater.
(16. 1.) than the interior and op-
posite angle ABC ; to each of
these add the angle ACB ; there*
fore the angles ACD, ACB are

greater than the angles ABC, "n
ACB ; but ACD, ACB are to- '^

f ether equal (13. 1.) to two right angles : therefore the angles ABC,
»CA are less than two right angles. In like manner, it may be de-
monstrated, that BAC, ACB, as also CAB, ABC, are less than two
right angles. Therefore, any two angles, &;c. Q. £• D.

PROP. XVin. THEOR.

The greater side of every triangle has the greater angle opposite to iU

Let ABC be a triangle of which,
the side AC is greater than the side
AB ; the angle ABC is also greater
than the angle BCA.

From AC, which is greater than
AB, cut off (3. 1.) AD equal to AB,
and join BD : and because ADB is
the exterior angle of the triangle
.fiDC, it is greater (16. 1.) than the
interior and opposite angle DCB ; but ADB is equal (5. 1.) to ABD,
because the side AB is equal to the side AD ; therefore the angle
ABD is likewise greater than the angle ACB ; wherefore much more
is the angle ABC greater than ACB. Therefore the greater side,
&;c. Q. E. D.

, I

OP GEOMETRY. BOOK I.

PROP. XIX. THEOR*

The greater angle of every triangle is subtended h^the greater 9ide^ er

nas the greater side opposite to it*

Let ABC be a triangle, of which the angle ABC is greatier than
the angle BCA ; the side AC is likewise greater than ttie $i4e AB^

For, if it be not greater, AC must
either be equal to AB, or less than it ;
it is not equal, because then the an-
gle ABC would be equal (5^ 1.) to the
angle ACB ; but it is not ; therefore
AC is not equal to AB ; neither is it
less ; because then the angle ABC
would be less (18. 1.) than the angle
ACB ; but it is not ; therefore the
side AC is not less than AB ; and it has been shewn that it is not eqvsj
to AB ; therefore AC is^ greater thm AB. Wherefoi^ the grealer
angle, &c. Q. E. D.

PROP. XX. THEOR.

[Any tftfo sides of a triangle are together greater than the third sidem

Let ABC be a triangle ; any two sides of it together are greater
than the third side, vis. the sides BA, AC greater than the side BC ;
and AB, BC greater than AC ; and BC, CA greater than AB.

Produce BA to the point D, ^

and make {3. 1.) AD equal to
AG ; and join DC. ly

Because DA is equal to AC, ^

the angle ADC is likewise equal
(5. 1.) to ACD ; but the angle
BCD is greater than the angle

ACD ; therefore the angle BCD -^ — — C

is greater than the angle ADC; ^
and because the angle BCD of the triangle DCB is greater than its
angle BDC,'and that the greater (19. 1.) side is opposite to the greater
angle : therefore the side DB is greater than tlxe side BC ; but DB
is equal to BA and AC together ; therefore B A and AC together are
greater than BC. In the same manner it may be demonstrated, that
the sides AB, BC arc greater than CA, and BC, CA greater thaa
AB. Therefore any two sides, &c. Q. E. D.

34 ELEMENTS

PROP. XXI. THEOR.

If from the ends of one side of a triangle, there he draum two straight
lines to a point within the iriangley these two lines shall be less thctn
the 9ther two sides of the triangle, hut shall contain a greater angle.

Let the two straight lines BD, CD he drawn from B, C, the ends
of the side BC of the triangle ABC, to the point D within it ; BD
and DC are less than the other. two sides BA, AC of the triangle, hut
contain an angle BDC greater than the angle BAC.

Produce BD to E ; and because two sides of a triangle (20. \l) are
greater than the third side, the two
Aides BA, AE of the triangle ABE
are greater than BE. To each of
these add EC ; therefore the sides
BA, AC are greater than BE, EC :
Again, because the two sides CE,
ED; of the triangle CED are greater
than CD, if DB be added to each,
the sides CE, EB, will be greater
than CD, DB ; but it has been shewn
that BA, AC are greater than BE,
EC ; much more then are BA, AC
greater than BD, DC.

Again, because the exterior angle of a triangle (16. L) is greater
than the interior and opposite angle, the exterior angle BDC of the
triangle CDE is greater than CED ; for the same reason, the exterior
angle CEB of the triangle ABE is greater than BAC ; and it has been
demonstrated that the angle BDC is greater, than the angle CEB ;
much more then is the angle BDC greater than the angle BAC.
Therefore, if from the ends of, 6dc, Q. £. D.

PROP. XXIL PROB.

To construct a triangle of which the sides shall he equal to three given
straight lines ; hut any two whatever of these lines must he greater
than the third {20. I.)

Let A, B, C be the three given straight lines, of which any two
whatever are greater than the third, viz. A an<i B greater than C ;
A and C greater than B ; and B and C than A. It is required to
make a triatigle of which the sides shall be equal to A, B> C, each to
each.

OF GEOMETRY. BOOK I.

Take a straight line DE;
terminated at the point D,
but unlimited towards E,
and make (3. 1.) DF equal
to A, FG to B, and GH
equal to C ; and from the
centre F, qt the distance
FD, describe (3. Post.)
the circle DKL ; and from
the centre G, at the dis-
tance GH, describe (3.
Post.) another circle HLK ;
and join KF, KG ; the tri-
angle KFG has its sides
equal to the three straight
lines, A, B, C.

Because the point F is the centre of th^ circle DKL, FD i» equal
(11. Def.) to FK ; but FD is equal to the straight line A ;• therefore
FK is equal to A : Again, because G is /the centre of the circle
LKH, GH is equal (11. Def.) to GK ; but GH is equal to C ; there-
fore, also, GK is equal to C ; and FG is equal to B ; therefore the
three straight lines KF, FG, GK, are equal to the three A, B, C :
And therefore the triangle KFG has its three sides KF, FG, GK
equal to the three given straight lines, A, B, C. Which was to be
done.

PROP. XXIII. PROB.

At a given point in a given straight line, to make a rectilineal angle

equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and
DCE the given rectilineal angle ; it is required to make an angle at
the given point A in the given
straight line AB, that shall be
equal to the given rectilineal
angle DCE.

Take in CD, CE any points
D, E, and join DE ; and make
(22. 1.) the triangle AFG,
the sides of which shall be
equal to- the three straight
lines, CD, DE, CE, so that
CD be equal to AF, CE to
AG, and DE to FG ; and be-
cause DC, CE are equal to
FA, AG, each to each, and the
base DE to the base FG ; the angle DCE is equal (8. 1.) to the angle

ELEMENTS

FAG. Therefore, at the given point A in the given straight line AB,
the angle FAG is made equal to the given rectilineal angle DCE*
Which was to be done.

PROP. XXIV. THEOR.

l^fiwo trumglea have two sides of the one equal to ttoo sides of the other^
each to eachy but the angle contained by tJie two sides of the one great--
er than the angle contained by the two sides of t?te other ; the base of
that which has the greater angle shaU be greater than the base of the
other.

Let ABC, DEF be two triangles which have the two sides AB, AG
equal to the two DE, DF each to each, viz. AB equal to DE, and
AC to DF ; but the angle BAC greater than the angle EDF ; the
base BC is also greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater
than the other, and at the point D, in the straight line DE, make
(23. 1.) the angle EDG equal to the angle BAC : and make DG equal
(3. L) to AC or DF, and join EG, GF.

Because AB is equal to DE, and AC to DG, the two sides BA, AC
are equal to the two ED, DG, each to each, and the angle BAC is
equal to the angle a T)

EDG, therefore the "^ '^

base BC is equal
(4. L) to the base
EG ; and because
DG is equal to DF,
the angle DFG is
equal (5. 1.) to the
angle DGF ; but the .
angle DGF is great-
er than the angle.
EOF ; therefore the
angle DFG is great-
er than EGF ; and much more is the angle EFG greater than the
angle EGF ; and because the angle EFG of the triangle EFG is
greater than its angle EGF, and because the greater (9. 1.) side is
opposite to the greater angle, the . side £0 is greater than the side
EF ; but EG is equal to BC ; and therefore also BC is greater than
EF. Therefore, if two triangles, dsc Q. E. D.

PROP. XXV. THEOR.

If two triangles have two sides of the one equal to two sides of the other,
each to each, but the base of the one greater than the base of the other ;
the angle contained by the sides of that which has the greater base,
shall be greater than the angle contained by the sides of the other.

Let ABC, DEF be two triangles which have the two sides^ AB, AC>

OF GEOMETRY. BOOK I.

equal to the two sides DE, DF, each to each, viz. AB equal to DE,
and AC to DF : but let the base CB be greater than the base EF,
the angle BAG is likewise greater than the angle EDF.

For, if it be not greater, it must either be equal to it, or less ; but
the angle BAG is not equal to the angle EDF, because then the base
BC would be equal (4. 1.) A 15

to EF ; but it is not ; there-
fore the angle BAG is not
equal to the angle EDF;
neither is it less ; because
then the base BC would be
less (24. 1 .) than the base
EF ; but it is not ; there-
fore the angle BAG is not
less than the angle EDF :
and it was shewn that it is -ri

not equal to it : therefore U ^ J*

the angle BAG is greater than the angle EDF. Wherefore, if two
triangles, &c. Q. E. D.

PROP. XXVL THEOR.

If tuDO triangles have two angles of the one equal to two angles of the

otlier, each to each ; and one side equal to one side, viz. either the

- side adjacent to the equal angles, or the sides opposite to the equdi

angles in each ; then shall the other side he equal, each to each ; and

also the third angle of the one to the third angle of the othet:*

Let ABC, DEF be two

triangles, which have the
angles ABG, BGA equal
to the angles DEF, EFD,
viz. ABG to DEF, and
BCA to EFD, also one
side equal to one side ;
and first, let those sides
be equal which are adja-
cent to the angles that
are equal in the two tri-
angles, viz. BC to EF ;
the other sides shall be equal, each to each, viz. AB to DE, and AG
to DF ; and the third angle BAG to the third angle EDF.

For, if AB be not equal to DE, one of them must be the greater.
Let AB be the grea;ter of the two, and make BG equal to DE, ancfjoin
GC ; therefore, because BG is equal to DE, and BG to EF, the two
sides GB, BG are equal to the two DE, EF, each to each ;. and the
angle GBC is equal to the angle DEF ; therefore 'the base GC is
equal (4. 1.) to the base DF, and the triangle GBC to the triangle

.38

ELEMENTS

DEF, and the other angles to the other angles, each to each, to which
the equal sides are opposite ; therefore the angle GCB is equal to
the angle DFE, but DFE is, by the hypothesis, equal to the angle
BCA ; wherefore also the angle BCG is equal to the angle BCA, the
less to the greater, which is impossible ; therefore AB is not unequal
to DE, that is, it is equal to it ; and BC is equal to EF ; therefore
the two AB, BC are equal to the two DE, EF, each to each ; and the
angle ABC is equal to the angle DEF ; therefore the base AC is
equal (4. 1.) to the base DF, and the angle BAG to the angle EDF.

Next, let the sides A - D

which are opposite to'
equal angles in each tri-.
angle be equal to one an-
other, viz. AB to DE ;
likewise in this case, the
other sides shall be equal,
AC to DF, and BC to
EF ; and also the third
angle BAC to the third
EDF. B H C E F

For, if BC be not equal to EF, let BC be the greater of them, and
make BH equal to EF, and join AH ; and because BH is equal ta
EF, and AB to DE ; the two. AB, BH are equal to the two DE, EF,
each to each ; and they contain equal angles ; therefore (4. 1.) the
base AH is equal to the base DF, and the triangle ABH to the trian-
gle DEF, and the other angles are equal, each to each, to which the
equal sfdes are opposite ; therefore the angle BEIA is equal to the an-
gle EFD ; but EFD is equal to the angle BCA ; therefore also the
angle BHA is equal to the angle BCA, that is, the exterior angle
BH A of the triangle AHC is equal to its interior and opposite angle
BCA, which is impossible (16. 1.) ; wherefore BC is not unequal to
EF, that is, it is equal to it ; and AB is equal to DE ; therefore the
two AB, BC are equal to the two DE, EF, each to each ; and they
contain equal angle? ; wherefore the base AC is equal to the base
DF, and the third angle BAC to the third angle EDF. Therefore^
if two triangles, &c. Q. E. D. .

PROP. XXVn. THEOR.

If a straight line falling upon two other straight lines makes the alternate
angles equal to one another^ these two straight lines are parallel.

Let the straight line EF, which falls upon the two straight lines AB»
CD make the alternate angles AEF, EFD equal to one another ; AB
is parallel to CD.

For, if it be not parallel, AB and CD being produced shall meet
either towards B, D, or towards A, C ; let them be produced and
meet towards B; D in the point 6 ; therefore GEF is a triangle, and

OP GEOMETRY* BOOK I.

its exterior angle AEF is greater (16. 1.) than the interior tiind oppo-
site angle EFG ; but it is
also equal Xo it, which is
impossible : therefore, AB
and CD b^ing produced,
do not meet towards B,
D. In like manner it may
be demonstrated that they
do not meet towards A,
C; but those straight lines

which meet neither way, though produced ever so far, are. parallel
(30. Def.) to one another. AB therefore is parallel to CD. Where-
fore, if a straight line, dz^c. Q« E. D.

PROP. XXVIII. THEOR*

If a straight line falling upon two other straight lines makes the exterior
angle equal to the interior and opposite upon the same side of the
line ; or makes the interior angles upon the^ same side together equal to
ttoo right angles; the tujo straight lines are parallel to one another.

Let the straight line EF, which falls upon the two straight lines
AB, CD, make the exterior angle EGB equal to GHD, the inte-
rior and opposite angle upon the
same side ; or let it make the
interior angles on the same
side BGH, GHD together equal
to two right angles ; AB is pa-
rallel to CD.

Because the angle EGB is
equal to the angle GHD, and
also (15. 1.) to the angle AGH,
the angle AGH is equal to the
angle GHD ; and they are the
alternate angles; therefore AB is parallel (27. 1.) to CD. Agaidj
because the angles BGH, GHD are equal (by Hyp.) to right an-
gles, and AGH, BGH, are also equal (13. 1.) to two right angles, the
angles AGH, BGH are equal to the angles BGH, GHD : Takeaway
the commonrangle BGH; therefore the remaining angle AGH is equal .
to the remaining angle GHD ; and they are alternate angles ; there-
fore AB is parallel to CD. Wherefore, if a straight Hne, &c.^ Q. E. D.. ^

PROP. XXIX. THEOR.

If a straight line fall upon two parallel straight lines^ it makes the alter'
note angles equal to one another ; and the exterior angle equal to the
interior and opposite upon the same side ; and likewise the two inte*
rior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight lines AB,
CD ; the alternate angles AGH, GHD are equal to one another ; and

40 ELEMENTS

the exterior angle BGB is equal to the interior and opposite, upon
the same side, GHD ; and the two interior angles BGH, 6HD upon
the same side are together equal to two right angles.

For if A^rH be not equaf to GHD, let KG be drawn making the
angle EGH equal to GHD, and produce KG to L ; then KL will be
parallel to CD (27. 1.) ; bat
AB is also parallel to CD ;
therefore two straight lines
are drawn through the same '
point G, parallel to CD, and
yet not coinciding with one
another, which is impossible
(1 1 . Ax.) The angles AGH,
GHD therefore are not un-

equal, that is, they are equal ^Tii

to one another. Now, the ^

angle EGB is equal to AGH (15. 1.) ; ahd AGH is proved to bo
equal to GHD ; therefore EGB is likewise equal to GHD ; add to
each of these the angle BGH; therefore the angles EGB, BGH are
equal to the angles BGH, GHD ; but EGB, BGH are equal (13. 1.)
to two right angles ; therefore also BGH, GHD are equal to two right
angles. Wherefore, if a straight line, dec. Q. E. D.

Cor. If two lines KL and CD make, with EF, the two angles
KGH, GHC together less than two right angles, KG and CH will
meet on the side of EF on which the two angles are tha^ are less than
two right angles.

For, if not, KL and CD are cither parallel, or they meet on tlio
other side of EF ; but they are not parallel ; for the angles KGH,
GHC would then be equal to two right ang^les. Neither do they meet
on the other side of EF ; for the angles LGH, GHD would then he
two angles of a triangle, and less than two right angles ; but this is
impossible ; for the four angles KGH, HGL, CHG, GHD are together
equal to four right angles (13. 1.) of which the two, KGH, CHG, are
by supposition less than two right angles ; therefore the other two,
HGL, GHD are greater than two right angles. Therefore, since
KL and CD are not parallel, and since they do not meet towards L
and D, they must meet if produced towards K and G. •

PROP. XXX. THEOR.

Straight lines which are parallel /o the same straight line are parallel

to one another.

Let AB, CD, be each of them parallel to EF ; AB is also parallel
to CD. •

Let the straight line GHK cut AB, EF, CD ; and because GHK
cuts the parallel straight lines AB, EF, the angle AGH is equal

OP GEOMETRY. BOOK L

(39v L) to the angle GHF. A-
gain, because the straight line
GK cuts the parallel straiglit
lines EP, CD, the angle GHF
is equal (29. 1.) to the angle
GKD : and it was shewn that
the angle AGK is equal to the
angle GHF; therefore also AGK
is equal to GKD ; and they are
alternate angles ;* therefore AB
is parallel (27. 10 to CD. Where-
fore straight lines,- &c. Q. E. D.

PROP. XXXI. PROR

To draw a straight line through a given point parallel to a gioen

nlraight line.

Let A be the given point, and BC the given straight line, it is re<

quired tb draw a straight line
through the point A, parallel
to the straight line BC.

In BC take any point D, and
join AD ; and at the point A^
in the straight line AD, make

~:ff

(23. 1.) the angle DAE equal to the angle ADC ; and produce the
straight line EA to F.

Because the straight line AD, which meets the two straight lines
BC, £F, makes the alternate angles EAD, ADC equal to one another^
£F is parallel (27. 1.) to BC. Therefore the strajight hne EAF is
drawn through the given point A parallel to the given straight lino
BC. Which was to be done.

PROP. XXXII. THEOR.

If a side of any triangle he produced^ the exterior angle is equal to the
two interior and opposite angles ; an£ the three interior angles of
every triangle are equod to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced to
D ; the exterior angle ACD is equal to the two interior and opposite
angles CAB, ABC ; and the three interior angles of the triangle, viz.
ABC) BCA, CAB, are together equal to two right angles.

Through the point C draw
GE parallel (31. 1.) to the
straightlineAB; and because
AB is parallel to CE, and AC
meets them, the alternate an-
£rlesBAC,ACE are equal (29.
1.) Again, because AB is

ELEMENTS

parallel to CE, and BD falls upon them, the exterior angle ECt) iV
equal to the interior and opposite angle ABC, but the angle AC£
was shewn to be equal to the angle BAG ; therefore the whole ex-
terior angle ACD is equal to the two interior and opposite angles
CAB, ABC ; to these angles add the angle ACB, and the angles
ACD, ACB are equal to the three angles CBA, BAC, ACB ; but the
angles ACD, ACB are equal (13. 1.) to two right angles ; therefore
also the angles CBA, BAC, ACB are equal to two right angles.
Wherefore, if a side of a triangle, <Sz;c. Q. E. D..

Cor. 1 . All the interior angles of any rectilineal 6gure are equal
to twice as many right angles as the figure has sides, wanting four
right angles.

For any rectilineal figure ABCDE can be divided into as many tri-
angles as the figure has sides, by drawing straight lines from a point
F within the figure to each of its angles. And, by the preceding
proposition, all the angles of these tri-
angles are equal to twice as many right
angles as there are triangles, that is, as
there are sides of the figure ; and the
same angles are equal to the angles of
the figure, together with the angles at
the point F, which is the common ver.
tex of the triangles ; that is, (2 Cor.
15. 1.) together with four right angles.

Therefore, twice as many right angles . t^

as the figure has sides, are equal to all -^ ^

the angles of the figure, together with four right angles, that is, the
angles of the figure are equal to twice as many right angles as the
figure has sides, wanting four.

CoR. 2. All the exterior angles of any rectilineal figure are toge-
ther equal to four right angles.

Because every interior
angle ABC, with its adja-
cent exterior ABD, is e-
qual (13. 1.) to two right
angles ; therefore all the
interior, together with all
the exterior angles of the
figure, are equal to twice
as many right angles as
there are sides of the fi-
gure ; that is, by the fore-
going corollary, they are
equal to all the interior
angles of the figure, to-
gether with four right angles ; therefore all the exterior angles are
equal to four right angles.

OF GEOMETRY. BOOK I.

PROP. XXXIII. THEOR. ,

TAc straight lines which join the extremities of two equal and parallel
straight lines, towards the same parts, are also themselves equal and
partdieL \

Let AB, CD, be equal and parallel straight lines, and joined towards
the same parts by the straight lines AC, BD ; AC, BD are also equal
and parallel. ' . .

Join BC ; and because AB is pa-
rallel to CD, and EC meets thenr,
the alternate angles ABC, BCD are
equal (29. 1.) ; and because AB is
equal to CD, and BC common to the

two triangles ABC, DCB, the two .

sides AB, BC are equal to the two C D

DC, CB ; and the angle ABC is equal to the angle BCD; therefore
the base AC is equal (4. 1.) to the base BD, and the triangle ABC to
the triangle BCD, and the other angles to the other angles (4. 1.)
each to each, to which the equal sides are opposite ; therefore the
angle ACB is equal to the angle CBD ; and because the straight line
' BC meets the two straight lines AC, BD, and makes the alternate
angles ACB, CBD equal to one another, AC is parallel (27. 1.) to
BD ; and it was shewn to be equal to it. Therefore straight lines,
&c. Q, E. D.

PROP. XXXIV. THEOR,

The opposite sides and angles of a parallelogram are equal to one ano^
ther, and the diameter bisects it ; that is, divides it into two equal
parts.

N. B. A Parallelogram is a foor-sided figure, of which (he opposite sides ore parallel ; and
the diameter is a straight lioe joining two of its opposite angles.

Let ACDB be a parallelogram, of which BC is a diameter ; the
opposite sides and angles of the figure are equal to one another; and
the diameter BC bisects it.

- Because AB is parallel to CD, and
BC meets them, the alternate angles
ABC, BCD are equal (29. 1.) to one
another ; and because AC is parallel
to BD, and BC meets them, the al.
ternate angles ACB, CBD are equal
(29. 1.) to one another ; wherefore C 33

the two triangles ABC, CBD have two angles ABC, BCA in one,
equal to two angles BCD, CBD in the other, each to each, and the
side BC, which is adjacent to these equal angles, common to the two
triatigles ; therefore their other sides are equal, each to each, and

ELEMENTS

the third angle of the one to the third angle of the other (26, 1.) ;
vu. the side AB to the side CD, and AG to BD, and the angle BAG
equal to the angle BDC. And because the angle ABC isequal to the
angle BCD, and the angle CBP to the angle ACB, the whole angle
ABD is equal to the whole angle ACD: And the angle BAG has
been shewn to be equal to the angle BDC : therefore the opposite sides
and angles of a parallelogram' are equal to one another; also, its dia-
Bieter bisects it ; for AB being equafto CD, and BG common, the two
AB, BG are equal to the two DC, CB, each to each ; now the angle
ABC is equal to the angle BCD ; therefore the triangle ABC isequal
(4. 1.) to the triangle BCD, and the diameter BC divides the parallel-
ogram ACDB into two equal parts* Therefore, &c. Q. E. D.

PROP. XXXV. THEOR.

Pardllelogfams upon the same base and between the same pardltels,

are equal ta one another*

(see the 2d AND 3d figures.)

Let the parallelograms ABCD, £BCP be upon the same base BC,.
and between tbe same parallels AF» BG ;. the parallelogram ABCD
is equal to the parallelogram EBC^.

If the sides AD, DF of the paral-
lelograms ABCD, DBCF opposite to
the base BC be terminated in the
same point D ; it is plain that each of
the parallelograms is double (34. 1.)
of the trian^e BDC ; and they are _________

therefore equal to one another. S O

But, if the sides AD, EF, opposite to the base BC of the parallel-
ograms ABCD, EBCF, be not terminated in the same point ; then,
because ABCD is a parallelogram, AD is equal (34. 1.) to BC ; for
the same reason EF is equal to BC ; wherefore AD is equal (1. Ax.)
to EF ; and DE is common ; therefore the whole, or the remainder,
AE is equal (2. or 3. Ax.) to the whole, or the remainder DF; now AB
is also equal to DC ; therefore the two E A, AB are equal to the twa

3? A Ts i> :f

B C B C

FD, DC, each to each ; but the exterior angle FDC is equal (29. 1.)
to the interior EAB, wherefore the base EB is equal to the base FC4

OF GEOMETRY. BOOK I.

and the triangle E AB (4. 1.) to the triangle FDC. Take the triangle
FDC from the trapezium ABGF, and from the same trapezium t^e
the triangle EAlf ; the remainders will then he equal (8. Ax.), that
is, the parallelogram ABCD is equal to the parallelogram EBGF.
Therefore, parallelograms upon the same hase, &c. Q. £. D.

• PROP. XXXVI. THEOR.

Parallelograms upon equal bases j and "between the some foraUdSy are

equal to one another*

•

Let ABCD, EFGH he parallelograms upon equal bases BC, FG,
and between the same paral-
lels AH, BG ; the parallelo-
gram ABCD is equal to
EFGH.

Join BE, CH; and be-
cause BC is equal to FG,
and FG to (34. 1.) EH, BC

is equal to EH ; and th6y -n C^ 1^ Ci

are parallels, and joined to- z*^ ' ^ ■*' ^

wards the same parts by the straight lines BE, CH : But straight lines
which join equal and parallel straight lines towards the same parts,
are themselves equal and parallel (33^^!.) ; therefore EB, CH are
both equal and parallel, and EBCH is a parallelogram ; and it is
equal (35. 1.) to ABCD, because it is upon the same base BC, and
between the same parallels BC, AH ; For the like reason, the paral.
lelogram EFGH is equal to the same EBCH : Therefore also the
parallelogram ABCD is equal to EFGH. Wherefore, parallelo«
grams, &c« Q. £. D.

PROP. XXXVn. THEOR.

Triangles upon the same base, and between the same parallels^ ateequdi

to one another.

Let the triangles ABC, DBC be upon the same base BC, and be*
'tween the same parallels, AD,

BC : The triangle ABC is E A. D !F

equal to ihe triangle DBC.

Produce AD both ways to the
points E, F, and through B draw
(31. 1.) BE parallel to CA ; and
through C draw CF parallel to

BD : Therefore, each of the .-5^ p

figures EBC A, DBCF is a paral- ^ ^

lelogram ; and DBC A is equal (35. 1.) to DBCF, because they are
upon the same base BC, and between the same parallels BC, £F ;

tLEMENTS

but the triangle ABC is the half of the parallelogram EBC A, because
the diameter AB bisects (34. 1.) it ; and the triangle DBC is the half
of the parallelogram DBCF, because the diameter DC bisects it; and
the halves of equal things are equal (7. Ax.) ; therefore the triangle
ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. i).

PROP. XXXVIII. THEOR. •

Driangles upon equal bases, and between the same parallels, are equal

to one another,

Let the triangles ABC, DEF be upon equal bases BC, £F, and
between the same parallels BF, AD : The triangle ABC is equal to
the triangle DEF.

Produce AD both Ways to the points G, H, and through B draw BG
parallel (31. 1.) to CA, and through F draw FH parallel to ED :
Then each of the figures ^ a -rv -rr

GBCA, DEFH is a paral- ^r A JJ ±L

lelogram ; and they are
equal *to (36. 1.) one an-
other, because they are
upon equal bases BC, EF,
and between the same pa-
rallels BF, GH ; and the r^ -wtk -m
triangle ABC is the half B C E B
(34. 1«) of the parallelogram GBCA, because the diameter AB bisects
it ; and the triangle DEF is the half (34. 1.) of the parallelogram
DEFH, because the diameter DF bisects it ; But the halves of equal
things are equal (7. Ax.) ; therefore the triangle ABC is equal to the
triangle DEF. Wherefore triangles, &c. Q. E. D.

PROP. XXXIX. THEOR.

Equal triangles upon the same base, and upon the same side of it, are

between the same parallels.

Let the equal triangles ABC, DBC be upon the same base BC, and
upon the same side of it; they are between the same parallels.

Join AD : AD is parallel to BC ; for, if it is not, through the point
A draw (31. 1.) AE parallel to BC, and join EC : The triangle ABC,
is equal (37. 1.) to the triangle EBC, because it is upon the same base
BC, and between the same parallels BC,
AE ; But the triangle ABC is equal to the A. D

triangle BDC ; therefore also the triangle
BDC is equal to the triangle EBC, the great-
er to the less, which is impossible : There-
fore AE is not parallel to BC. In the same
manner, it may be demonstrated that no
other line but AD is parallel to BC ; AD
is therefore parallel to it. Wherefore equal
triangles upon^ dfc» Q. E. D.

OF GEOMETRY. BOOK I. 47

PROP. XL. THEOR.

Equal triangles on Ike same sidM of hoses uihich are equal and in the
same straight line, are between the same parallels.

Let the equal triangles ABC, DEF be upon equal bases EC, EF,
in the same straight line BF, a^ JJ

and towards the same parts ;
they are between the same pa-
rallels.

Join AD ; AD is parallel to
'BC ; For, if it is not, through A

draw (31. 1.) AG parallel to _ r^ tp xh

BF, and join GF. The triangle j3 ^ JU If

ABC is equal (38. L) to the triangle GEF, because they are upon
equal bases BC, EF, and between the same parallels BF, AG ; But
the triangle ABC is equal to the triangle DEF ; therefore also the
triangle DEF is equal to the triangle GEF, the greater to the less,
which is impossible ; Therefore AG is not parallel to BF ; and in the
same manner it may be demonstrated that there is no other parallel
to it but AD ; AD is therefore parallel to BF. Wherefore equal tri-
angles, &c. Q. E. D.

PROP. XLI. * THEOR.

J^ a 'parallelogram and a triangle he upon the same hasey and between
the same parallel ; the parallelogram is double of the triangle.

Let the parallelogram ABCD and the triangle EEC be upon the
same base BC and between the same pa-
rallels BC,AE; the parallelogram ABCD A D E
is double of the triangle EBC.

Join AC ; then the' triangle A6C is equal
(37. L) to the triangle EBC, because they
are upon the same base BC, and between
the same parallels BC, AE. But the pa-
rallelogram ABCD is double (34. 1^.) of
the triangle ABC, because the diameter
AG divides it into two equal parts ; where-
fore ABCD is also double of the triangle
EBC. Therefore, if a parallelogram, <&c.

PROP. XLIL PROB.

To describe a parallelogram that shall be equal to a given triangle, and
have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle«

ELfiMENTS

It is required to describe a parallelogram that jsihall be equal to the
given triangle ABC, and have one of its angles equal to D.

Bisect (10. 1.) EC in £, join a£, and at the point E in the straight
line EC make (23. 1 .) the angle CEF equal to D ; and through A
draw (31. 1.) AG parallel to BG, and through a draw CG (31. 1.)
parallel to EF ; Therefore FECG A "P O

is a parallelogram : And because " ^ | i j

BE is equal to EC, the triangle ^v [

ABE is likewise equal (38. 1.) to
the triangle AEC, since they are
upon equal bases BEf EC, and
between the same parallels BC
AG ; therefore the triangle ABC
is double of the triangle AEC.
And the parallelogram FECG is
likewise double (41. 1.) of the
triangle AEC, because it is upon the same base, and between the
same parallels : Therefore the parallelogram FECG is equal to the
triangle ABC, and it has one of its angles CEF equal to the given
angle D ; Wherefore there has been described a parallelogram
FECG equal, to a given triangle ABC, having one of its angles CEF
equal to the given angle D. Which was to be done.

PROP. XLIII. THEQR.

The camplemenls of the parallelograms which are about the diameter of
any parallelogram, are equal to one another.

Let ABCD be a parallelogram of which the diameter is AC ; let
EH, FG be the parallelograms about AC, that is, through which AG
passes, and let BK, KD be the other a -rr -rv

parallelograms,, which make up the -^ ■"- -*^

whole figure ABCD, and are there-

fore called the complements ; The "P / Ni^-'V- JTR*

complement BK is equal to the com-
plement KD.

Because ABCD is a parallelogram
and AC its diameter, the triangle
ABC is equal (34. 1.) to the triangle
ADC ; And because EKHA is a pa- [Q
rallelogram, and AK its diameter, the
triangle AEK is equal tathe triangle AHK : For the same reason, the
triangle KGC is equal to the triangle KFC. Then, because the tri-
angle AEK is equal to the triangle AHK, and the triangle KGC to
the triangle KFC ; the triangle AEK, together with the triangle KGC,
is equal to the triangle AHK, together with the triangle KFC : But
the whole triangle ABC is equal to the whole ADC ; therefore the re-
maining complement BK is equal to the remaining complement KD.
Wherefore, the complements, <&c. Q. E. D.

OPGEOMfiTRT. BOOK I.

PROP. XLIV. PROB.

To a given ttraigkt lintio apply a paraUdogramtmkieh MU tecfiMl
to a given iriangle^ and have one ofiU angke e^pud to a given recti'
Uneal angle*

Let AB be the given straight line, and C the given triangle, and D
the given rectilineal angle. It is required to appl j to the straight line
AB a perallelograni equal to the triangle C, and having an angle equal
lo D. Make (42. 1.) the parallelogram BEF6 equed to the triangle

G, having the angle £B6 equal to the angle D, and the side BE in the
same straight line w*^ AB : produce FG to H, and through A draw
(31. 1.) AH par^ to B6 or EF, and join HB. Then because the
straight line HP f .lis upon the parallels AH, £F, the angles AHF,
HF£y are together equal (29. 1«) to two right angles ; wherefore the
angles BHF, HFE are less than two right angles ; But straight lines
which with another straight line make the interior angles, upon the
eame side less than two right angles, do meet if produceo (Cor. 29. l.)t
Therefore HB, FE will meet, if produced ; let them meet in K, and
through K draw KL parallel to EA or FH, and produce HA, GB to
the points L, M : Then HLKF is a parallelogram, of which the diame*
ter is HK, jand AG, ME are the parallelograms about HK ; and LB}
BF are the complements ; therefore LB is equal (43. 1.) to BF : but
BF IS equal to the triangle C ; wherefore LB is equal to the triangle
C ; and because the angle GBE is equal (15. 1.) to the angle ABM,
and likewise to the angle D ; the angle ABM is equal to the angle I>:
Therefore the parallelogram LB, which is applied to the straight line
AB, is equal to the triuigle C, and has the angle ABM equcd to the
angle D^: Which' was to be done.

PROP. XLV, PROB.

To describe a parallelogram equal to a given rectilineal jigure^ and Aao-
ing an an^ equal to a given rectilineal angle*

Let A BCD be (he given rectilineal figure, and E the given rectili*
neal angle. It is required to describe a parallelogram equ|il to ABCD,
and having an angle equal to E.

iO ELEMENTS

Join DB, and describe (42, 1.) the panllelogram FH equal to the
triangle ADB, and having the angle HKF equal to the angle E ; and to
tlie straight line 6H (44. 1.) apply the parallelogram GM equal to the
triangle DBC, having the angle GHM equal to the angle E. And be-
cause the angle E is equal to each of the angles FKH, GHM, the angle
FKH is equal to GHM ; add to each of these the angle KHG ; there,
fere the angles ^KH, KHG are equal to the angles KHG, GHM ;
but FKH, kHG are equal (29. 1.) to two right angles ; therefore also
KHG, GHM are equal to two right angles : and because at the point

A B B: H • M

H in the straight line GH, the two straight lines KH, HM, upon the
opposite sides of GH, make the adjacent angles equal to two right ao*
gles, KH is in the same straight line (14. 1.) with HM. And be*
dause the straight line HG meets the parallels KM, FG, the alterttate
angles MHG, HGF are equal (29. 1.) ; add to each of these the an-
gle HGL ; therefore the angles MHG, HGL, are equal tb the angles
HGF, HGL: But the angles MHG, HGL, are equal (29. 1.) to twe
right angles ; wherefore also the angles HGF, HGL, are equal to two.
right angles, and FG is therefore in the same straight line with GL.
And because KF is parallel to HG, and HG to ML, KF is parallel (30.
1.) to ML ; but KM, FL are parallels : wherefore KFLM isa parallel-
ogram. And because the triangle ABD is equal to the parallelogram
HF, and the triangle DBC to the parallelogram GM, the whole rec-
tilineal figure ABOD is dqual to the whole parallelogram KFLM ;
therefore the parallelogram KFLM has been described equal to the
given rectilineal figure ABGD, having the angle FKM equal to the
given angle E. Which was to be done.

Cor. From this it is manifest how to a given straight line to apply
a parallelogram, which shall have an angle equal to a given rectili-
neal angle, and shall be equal to a given rectilineal figure, viz, by
applying (44. 1.) to the given straight line a parallelogram equal to
the first triangle ABD, and having an angle .equal to the given angle.

PROP. XLVI. PROB.

To describe a square upon a given straight line.

Let AB be the given straight line : it is required to describe a
square upon AB.

From the point A draw (11. 1.) AC at right angles to AB ; aad
make (3. 1.) AD equal to AB, andthrough the point D draw DE pa-

OF GEOMETRY. BOOK I.

Tallel (31. 1.) to AB, and through B draw BE parallel to AD ; there-
fore A DEB is a parallelogram ; whence AB is equal (34. 1.) to DE,
and AD to BE ; but BA is equal to AD - Q
therefore the four straight lines BA, AD,
DE, EB are equal to one another, and
the parallelogram ADEB is equilateral ;
it is likewise rectangular ; for the straight
line AD meeting the parallels, AB, D£,
makes the angles BAD, ADE equal (29.
1.) to two right angles ; but BAD is aright
angle ; therefore also ADE is a right angle
now the opposite angles of parallelograms
are equal (34. 1.) ; therefore each of the
opposite angles ABE, BED is a right an-
gle ; wherefore the figure ADEB is rect-
angular, and it has been demonstrated that it is equilateral ; it is
therefore a square, and it is described upon the given straight line
' AB ; Which was to be done.

Cor. Hence every parallelogram that has one right angle has all
Its angles right angles.

PROP. XLVII. THEOR.

in any right angled triangle, the square vMch is described upon the
side subtending the right ar^le, is equal to the squares described up-
on the sides which contain the right angle.

Let ABC be a right angled triangle having the right angle BAC ;
the Siquare described upon the side EC is equal to the squares des-
cribed upon BA, AC.

On BC describe (46. 1.) the square BDEC, and on BA, AC the
squares GB, HG ; and through A draw (31. 1.) AL parallel to BD or
CE, and join AD, FC ;wthen, because each of the angles BAC, B AO
is a right angle (25. def.), the
two straight lines AC, AG upon
the opposite sides of AB, make
with it at the point A the adja-
cent angles equal to two right an-
gles ; therefore CA is in the same
straight line (14. 1.) with AG ;
for the same reason, AB and AH
are in the same straight line.
Now because the angle DBC is
equal to the angle FBA, each of
them being a right angle, adding
to each the angle ABCj the whole
angle DBA will be equal (2. Ax.)*
to the whole FBC; and because
the two sides AB, BD, are equal
(o the two FB, BC each to each,
«nd the angl^DBA equal to the

52 ELEHKNTS, 6ce.

angle FBC, therefore the base AD is equal (4. 1 .) (a the fmee FC,
and the triangle ABD to the triangle FEC. But the parallelogran*
BL is double (41. 1.) of the triangle ABD, because they are upon
the same base, BD, and between the same parallels, BD, AL; and the
aquare GB is doable of the triangle BFC beeause these also are upon
the same base FB, and between the same parallels FB, 6G. Now
the doubles of equals are equal (6. Ax.) to one another ; therefore
the parallelogram BL is equal to the square GB : And in the same
maanor, by joming AE, BK» it is demonstrated that the parallelogram
CL is equal to the square HC. Therefore, the whole square BDEC
is equal to the two squares GB, HC ; and the square BDEC is de*
scribed upon the straight line BC, and the squares GB, HC upon BAr
AC : wherefore the square upon the side BC is equal to tbe squares
upon the sides BA^ AC. Therefore, in any right angled triangle, 6ic^
Q. E. D.

PROP. XLVlII. THEOR.

J^the square described upon tme of the sides of a triangle^ he eqmdto
the squares described t^pon the other ttoo sides of it ; the angle con-
tained by these two sidesis a right angle.

If the square described npon BC, one of the sides of tbe triangle
ABC, be equal to the squares upon the other sides BA, AC, the angle
BAC is a right angle.

From the point A draw (11. 1.) AD at right angles to AC, and
make AD equal to BA, and join DC. Then because DA is equal to
AB, the square of DA is equal to the square
of AB ; To each of these add the square of
AC ; therefore the squares of DA, AC are^
equal to the squares of BA, AC. But the
square of DC is equal to (47. 1.) to the squares
of DA, AC, because DAC is a right angle ;
and the square of BC, by hypothesis, is equal
to the squares of BA, AC ; therefore, the
square of DC is equal to the square of BC }
and therefore also the side DC is e^ual to the
aide BO. And because the side DA is equal to AB, and AC common
to the two triangles DAC, BAC, and the base DC likewise equal to
the base BC, the angle DAC is eq^al (6. 1.) to the angle BAC ; But
DAC is a right angle ; therefore also BAC is a right angle* There*
fore» if the square, &c. Q. E. D.

ELEMENTS

GEOMETRY,

BOOK II.

DEFINITIONS.
I.

EvEHT right angled parallelogram, or rectangle^ is said to be con-
tained by any two of the straight lines which are about one of the
right angles.

** Thus the right angled parallelogram AC is called the rectangle
*' contained by AD and DC, or by AD and AB, dec. For the sake of
** brevity, instead of the rectangle conlained by AD and DC, we shall
" simply say the rectangle AD. DC, placing a point between the two
** sides of the rectangle. Also, instead of the square of a line, for
" instance, of AD, we shall frequently in what foAows write AD'."

" The sign -f* placed between the names of two magnitudes, sig-
'' nifies that those magnitudes are to be added together ; and the sign
** — placed between them, signifies that the latter is to be taken away
^* from the former."

*' The sign = signifies, that the things between which it is placed
'' are equal to one another."

II.

In every parallelogram, any, of-
the parallelograms about a di-
ameter, together with the two
complements, is called a Gno-
mon. " Thus the paralleled-
" gram H6, together with the
<< complements AF, FC, is the
*' gnomon of the parallelogram
'*AC. This gnomon may also,
" for the sake of brevity, be
^' called the gnomon AGK or
«EHC.*'

ELEMENTS

PROP. I. THEOR.

if there he two Hrtdght lines^ one of which ie divided into any number
of parts ; the rectangle contained by the two straight lines is equal to
the rectangles contained by the undivided line^ and the several farts
of the divided line.

Let A and BC be two straight lines'; and let BC be divided into any
parts in the points D, E ; the rectangle A.BC is equal to the several
rectangles A.BD, A.DE, A;EC.

From the point B draw (11. 1.)
BF at right angles to BC, and
make BG equal (3. 1.) to A ; and
through G draw (31. 1.) GH pa-
rallel to BC ; and through D, E, C,
draw (31. I.) DK, EL, CH paral-
lel to BG ; then BH, BK, DL, and
£H are rectangles, and BH = BK
+DL+EH.

But BH=B6.BC=A.BC, be-
cause BG =: A : Also BK = BG.
BD = A.BD, because BG = A ;
and DL=DK.DE=A.DE, because (34. 1.) DK=BG=A7 In
like manner, EH = A.EC. Therefore A.BC = A.BD + A.DE +
A. EC ; that is, the rectangle A.BC is equal to the several rectangles
A.BD, A.DE, A.EC. Therefore, if there be two straight lines, &c.
Q. E. D. ' ' .

»«

PROP. XL THEOR.

Jf a straight line be divided into any two parts ^ the rectangles contairted
by the whole and each of the parts ^ are together equal to the square
of the whole line.

Let the straight line AB be divided into any two parts in the point
C ; the rectangle AB.BC, together with the a
rectangle AB.AC, is equal to the square of
AB; or AB.AC+AB.BC=AB=^.

On AB describe (46. 1.) the square ADEB,
and through draw CF (31. 1.) parallel to
AD or BE; then AF+CE=AE. But AF
= AD. AC = AB.AC, because AD =* AB ;
CE = BE.BC = AB.BC ; and AE = AB^
Therefore AB.AC+AB.BC=ABl There-
fore, if a straight line, d£C« Q. E. D. D

■'»;'■■

OF GEOMETRY. BOOK 11.

PROP. III. THEOR.

tf a straight line he divided into any two parts j the rectangle contained
by the whole and one of the partSy is equal to the rectangle contained
by the two parts^ together with the square of the aforesaid pari*

Let the straight line A6 be divided into two parts in the point O ;
the rectangle AB.BC is equal to the rectangle, AG.BG, together with

Upon BC describe (46. 1.) the ,
square CDEB, and produce ED to
F, and through A draw (31. 1.) AF
parallel to CD or BE ; then AE=
AD+CE.

But AE = AB.BE = AB.BC,
because BE=BC. So also AD=
AC.CD = AC.CB ; and CE =
BCS- therefore AB.BC = AC.CB
+BC^. Therefore, if a straight
liAe^ &c. Q. E. D.

PROP. IV. THEOR.

If a straight line be divided into any twofiartSy the square of the whole
line is equal to the squares of thetv30 partSy together with twice the
rectangle contained by the parts. ^ .

Let the straight line AB be divided into any two parts in C ; the
square of AB is equal to the squares of AC, CB, and to twice the rect-
angle contained by AC, CB, that is, AB^=AC2+OB^+2AC.CB.

Upon AB describe (46. 1.) the square ADEB, and join BD, and
through C draw (31. 1.) CGF parallel to AD or BE, and through G
draw HK parallel to AB or DE. And because CF is parallel to AD,
and BD falls upon them, the exterior an-
gle BGC is equal (29. 1.) to the interior
and opposite angle ADB ; but ADB is
equal (5. 1.) to the angle ABD, because
BA is equal to AD, being sides of a square ;
wherefore the angle GGB is equal to the
angle GBC ; and therefore the side BC is
equal (6. 1.) to the side CG'f but CB is
equal (34. 1.) also to GK and CG to BK :
wherefore the figure CGKB is equilateral.
It is likewise rectangular ; for the angle
CBK being a right anglo^the other angles of the parallelogram CGK^
are also right angles (Cor. 46. 1.) Wherefore CGKB is a square, and
it is upon the side CB. For the same reason HF also is a squarcy and

ELEMENTS

it is np6n the side HG, which is equal to AC : therefore HF, CR are
the squares of AC, CB. And because the complement AQ is equal
(43. 1.) to the complement GE'; and because AG= AC.CG=sAC.CB,
therefore also GE^AC.CB, and AG+GE=2AC.CB. Now, HF=3
AC" and CK=CB«; therefore, HF+CK+AG+GE=AC"+CB>+
2AC.GB.

But HF+CK+ AG +GE= the figure AE, or AB' ; therefore AB«
sAC'+CB'+SACCa Wherefore, if a straight line be divided,
dec. Q. E. D.

Cor. From the demonstration, it is manifest that the parallelograms
about the diameter of a square are likewise squares.

PROP. V. THEOR.

If a straight line be divided into two equal partSf and also into two un-
equal parts ; the rectangle contained by the unequal parts^ together
with the square of the line betvoeen the points' of section^ is equal to the
square of half the line*

Let the straight line AB be divided into two equal parts in the point

C, and into two unequal parts in the point D ; the rectangle AD.DB,

together with the square of CD, is equal to the square of CB, or AD.

DB+Cp"=CB".

UponCB describe (46. 1.) the square CEFB, join BE, and through

D draw (31. 1.) DHG parallel to CE or BF ; and through H draw

KLM parallel to CB or EF ; and* G

also through A dr^w AK parallel A^

to CL or BIVI : And because CH

=HF, if DM be added to both, jr

CM=DR ButAL=(36.1.)CM,

therefore AL=DF, and adding

CH to both, AH=gnomon CMG.

But AH = AD.DH = AD.DB,

because DH=DB (Cor. 4. 2.) ;

therefore gnomon CMG=AD.DB. To each add LG=CD^ th-efi,

gnomon CMG + LG = AD.DB + CD^ But CMG + LG = BC* ;

therefore AD.DB+CD'=BC^ Wherefore, if a straight line, <kc.

Q. E. D.

*' Cor. From, this proposition it is manifest, that the difference of
the squares of two unequal lines, AC, CD, is equal to the rectangle
contained by their sum and difference, or that AC^ — CD'=(AC+

« CD) (AC—CD)."

PROP. VI. THEOR.

If a straight line be bisected^ and produced to any point ; the rectangle
contained by the whole line thus produced, and the part of it produced^
together with the square of half the line bisected^ is equal to the square
of the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point

OP GEOMETRY. BOOK 11.

I) ) the rectangle AD»DB together with the sqUare of CB, is equal
to the square of CD.

Upon CD describe (46. 1.) the square CEFD, join DE, and through
fi draw (31 . 1 BHG parallel to CE or DF, and through H draw KLM
parallel to AD or EF, and also through A draw AK parallel to CL or
DM. And because AG is equal to OB, the rectangle AL is equal (36. 1 .)
to CH ; but CH is equal (43. 1.)
to HF ) therefore also AL is equal
to HF : To each of these add
CM ; therefore the whole AM is
equal to the gnomon CMG. Now
AM=AD.DM=AD.DB,because
DM = DB. Therefore gnomon
€MG=AD.DB, and CMG+LG
= AD.DB+CB^ But CMG +
LG = OF = CD^ therefore AD.
DB + CB^ + CD^ Therefore,
if a straight line, &c. Q. E. D.

PROP. VII. THEOR.

If a slraight line he divided into any ttco parts, the squares of t1^ whole
line, and of one of the parts, are equal to twice the rectangle contain-
ed by the whole and that part, together with the square of the othetpart*

Let the straight line AB be divided into any two parts in the point
C ; the squares of AB, BC, are equal to twice the rectangle AB.BC,
together with the square of AC, or AB''+BC^=2AB.BC+AG».
^ Upon AB describe (46. 1*) the square ADE^, and construet the
figure as in the preceding propositions: Because AG s^GE, AG+CK
:=GE+CK, that is, AK=CE, and there- ^ ^ ^

fore AK+CE ;= 2AK. But AK+CE= A U B

gnomon AKF+CK ; and therefore AKF
+CK=2AK=^2AB.BK = 2AB.BC, be.
cause BK=(Cor. 4. 2.) BC, Since then, JJ
AKF+CK=2AB.BC, AKF+CK+HF
=i2AB.BC+HF; and because AKF+
HF=AE=AB^ AB^+CK = 2AB.BC+
HF, that is, (since CK=*CB^ and HF
=AC^) AB^ + CB» = 2AB.BC + AC^
Wherefore, if a straight line, &c. Q. E. D. X)

Otherwise,

« Because AB^ =« AC + BC» + 2AC.BC (4. 2.) adding BC» to
« both, AB^+BC» =• ACH2BC»+2AC.BC.
"But BC^+AC.BC=AB.BC (3.2.); and
"therefore,2BC24-2AC.BC=2AB.BC;and ^

** therefore ABHBC»=AC2+2AB.BC." ^ ! ^

ELEMENTS

* Cor. Hence, the sum of the squares of any two lines is equaf
« to twice the rectangle contained by the lines together with the
<< square of the difference of the lines.''

PROP. VIII. THEOR.
y a itraighi line he divided into any two parts y four times the rectangle
contained hy the whole liney and one of the parts, together vnth the
square of the other party is equal to the square of the straight line
which is made up of the whole and the first-mentioned part.

Let the straight line AB be divided into any two parts in the point
C ; four times the rectangle AB.BC, together with the square of AC,
is equd to the square of the straight line made up of AB and BC to-
gether.

Produce AB to D, so that BD be equal to CB, and upon AD de-
scribe the square AEFD ; and construct two figures such as in the
preceding. Because GK is equal (34. 1.) to CB, and CB to BD, and
BDto KN,GK is equal 'to KN. For the same reason, PR is equal to
RO ; and because CB is equal to BD, and GK to KN, the rectangles
CK and BN are equal, as also the rectangles GR and RN : But CK
is equal (43. 1.) tojElN, because they are the complements of the
pan^llelogram CO : therefore also BN is equal to GR ; and the four
rectangles BN, CK, GR, RN are therefore equal to one another, and
so CK+BN+GR+RN=4CK. Again, because CB is equal to BI>,
and BD equal (Cor. 4. 2.) to BK, that ^ ^

is, to CG ; and CB equal to GK, that -^ -^

(Cor. 21. 2.) is, to GP ; therefore CG I

is equal to GP ; and because CG is ^ ,

equal to GP, and PR to RO, the rect- "^

angle AG is equal to MP, and PL to

RF : but MP is equal (43. 1.) to PL,^

because they are the complements of

the parallelogram ML ; wherefore

AG is equal also to RF. Therefore

the four rectangles AG, MP, PL, RF,

are equal to one another, and so AG

+MP+PL+RF=x4AG. And it was

demonstrated, that CK+BN+GR+RN=4CK ; wherefore, adding

equals to equals, the whole gnomon A0H=4AK. Now AK=AB.BK

=AB.BC,and4AK==^4AB3C ; therefore, gnomon AOH=4AB.BC;

and adding XH, or (Cor. 4. 2.) AC^ to both, gnomon AOH+XH=

4AB.BC+AC". But AOH+XH=AF=Al)^ therefore AD^=;=

4AB.BC+AC^ Now AD is the line that is made up of AB and BC,

added together into one line : Wherefore, if a straight line, &P.

Q* E. D.

" Cor. 1 . Hence, because AD is the sum» and AC the difference of
" the lines AB arid BC, four times the rectangle contained by any twa
« lines, together with the square of their difference, is equal to the
" square of the sum of the lines."

% -

•• •

OP GEOMETRY. BOOK 11. 69

^ Cor. 2. Prom the demonstration it is manifest, that since the-
*• square of CD is quadruple of the square of CB, the square of any
^< line is quadruple of the square of half that line."

Otherwise*; ,

«« Because AD is divided any how in G (4. 2.), AD»=AC2+CD»
« + 2CD.AC. But CD=2CB ; and therefore CD=GB^+BD'+
«' 2CB.BD (4. 2.)=4CB^ and alfio 2CD.AC=4CB.AC ; therefore,
« AD^ = AC +4BC* + 4BC. AC. Now BC'+BC.AC =s AB.BC
« (3. 2.) and therefore AD»=AC+4AB.BC. Q. E. D."

PROP. IX. THEOR.

If a straight line be divided into two equals and aUo into tteo unequal
parts; thesquarestf tJie two unequal p(urts are U)gether double (^ the
.square of hay^ the liney and of the square of the line between the points
<f section.

Let the straight line AB be divided at the point C into two equal,
and at D into two unequal parts ; The squares of AD, DB are toge-
ther double of the squares AC, CD.

From the point G draw (11.1.) CE at right angles to AB, and make
it equal to AG or GB, and join EA,EB ; through D draw (31. 1.) DF
parallel to CE, and through S draw FG parallel to AB ; and join AF;
Then, because AC is equal to CE, the angle E AC is equal (5. 1.) to the
angle AEG ; and because the angle AGE is a right angle, the two
others AEC, EAC together make one right angle (32. 1.) ; and they
are equal to one another ; each of them therefore is half of a right
angle. For the same reason each of _ JE

the angles GEB, EBG is half a right an-
gle ; and therefore the whole AEB is
a right angle ; And because the angle
GEF is half a right angle, and EGF a
right angle, for it is equal (29. 1.) to ^_
the interior and opposite angle ECB, j^ O 3C) B

the remaining angle EFG is half a right

angle ; therefone the angle GEF is equal to the angle EFG, and tbe
side EG equal (6. 1.) to the side GF ; Again, because the^nffl^ at b la
half a right angle, and FDB a right angle, for it is equal (i«J. 1.) to the
interior W opposite angle ECB, the remaining angle BFD is half a
right angle ; therefore the angle at i5 is e
side DF to (6/ 1.) the sideDB. Now, I

=2GP. But £^=£0" + GF* ; therefore. EP = aGP =2CD-,
^acause (34. 1.) CD=GP. And it was shown that AE? = 2AC j
therefore AE» -f EP=2AC* + 2Ciy. But (47. 1.) AP»=AE» +

60 ELEMENTS

EF«, and AD»+DP=AF«, or AD»+DB»=AP ; therefore, aW^
AD"+DB« = 2AC* + 2CD*. Therefore, if a straight line, fcc.
a E. D.

Otherwise ;

" Because AD«=(4. 2.) AC»+CD'+2AC.CD, and DB«+2Ba
•* CD=(7. 2.) BC»+CD»=AC*+CD", by adding equals to equals^
« AD«+BD'+2BC.CD=2AC«+2CD»+2AC.CD ; and therefore
'< taking away the equal rectangles 2BC.CD and 2AC.CD, there re-
« mains AD^+DB«=2AC«+2CD«."

PROP. X. THEQR.

If a straight line hehisecledy and pt'oduced to any point, the square of
the whde line thus^produced, txnd the square of the part of it produc-
edy are together double of the square of half the line hiseeted, and of
the square of the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point
D ; the squares of AD, DB are double of the squares of AC, CD.

From the point C draw (11.1.) CE at right angles to AB, and make-
it equal to AC or CB ;.join AE, EB ; through E draw (31. 1.) EF pa-
rallel to AB, and through D draw DF parallel to CE. . And because
the straight line EF meets the parallels EC, FD, the angles CEF, EFD
are equal (29. 1.) to two right angles ; and therefore the angles BEF,,
EFD are less than two right angles ; But straight lines, which with
another straight line make the interior angles upon the same side les»
than two right angles, do meet, (Cor. 29. 1.) if produced far enough ;
therefore £B, FD will meet, if produced, towards B, D : let thenk
meet in G, and join AG. Then because AC is equal to CE, the angle
CEA is equal (5. 1.) to the angle EAC ; and the angle ACE is a right
angle ; therefore each of the angles CEA, EAC is half a right angle
(32. 1.) ; For the same reason, each of the anglesCEBjEBCis half
a right angle ; therefore AEB is a right angle ; And because EEC i»
half a riglktangle, DfiGis also (15. 1.) half a right angle, for they are
vertically opposite : but BDG is a right angle, because it is equal (29.
1.) to the alternate angle DCE ; therefore the remaining angle PGB
is half a right angle, and is therefore equal to the angle DBG;, where-
fore also the side DB is equal (6. 1.) to the side DG. Again, because
EOF is half a right angle, and the angle at Fa right angle, being equal
(34. 1.) to the opposite angle toi -im

ECD, the reiaaining angle FEG ^^ ^

is half a right angle^ and equal to
the angle EGF ; wherefore also
the side GF iis equal (6. 1.) to
the side FE. And because EC
= CA, EC + CA« = 2CA\ /U
Now AE»= (47. 1.) AC^+CE;
therefore, AE'82AC^ Again,

OF OEOMETRY. BOOK II.

because EF=FG, EP=FG^ and EF«+FG»=2EP. But EG»=
(47. 1.) EP+FG^ ; therefore EG>=2EP ; and since EF=CD,
EG^=«=2CD^ And it was demonstrated, that AE"=2AC» ; therefore,
AE« + EG' = 2AC> +2CD». - Now, A6«=AE»+ EG", wherefore
AG3=2AC^+2CD». But AG« (47. I.) =AD»+DG«=AD'+DB%
because DG=DB : Therefore, AD'+DB«=:2AC'+2CD». Where,
fore, if a straight'line, &c. Q. E. D.

PROP. XI. PROB.

To divide a given straight line into ttoopairtSf $o that the rectangle con*
iained by the whole, and one of the parts f may be equai to the square
of the other part.

Let AB be the given straight line ; it is required to divide it into
two parts, so that the rectangle contained by the whole, and one of
the parts, shall be equal to the square of the other part;

Upon AB describe (46. 1.) the square ABDC ; bisect (10. 1.) AC
in £, and join BE ; produce CA to F, and make (3. 1.) EF equal to
EB, and upon AF describe (46. 1.) the square FGHA, AB is divided
in H, so that the rectangle AB, BH is equal to the square of AH.

Produce GH to K : Because the straight line AC is bisected in E,
and produced to the point F, the rectangle CF.FA, together with the
square of AE, is equal (6. 2.) to the square of EF : But EF is equal
to EB ; therefore the rectangle CF.FA, together with the square <^
AE, is equal to the square of EB ; And the squares of BA, A£ are
equal (47. 1.^ to the square of EB, be- ]□« a

cause the angle EAB is a right angle ;
therefore the rectangle CF.FA, toge-
ther with ihjb square of AE, is equal to
the squares of BA, AE : take away the
square of AE, which is common to both,
therefore the remaining rectangle CF.FA ^
is equal to the square of AB. Now the
figure FK is the rectangle CF.FA, for AF
is equal to FG ; and AD is the square of
AB ; therefore FK is equal to AD : take
away the common part AK, and the re- Bi
mainder FH is equal to the remainder
HD. But HD is the rectangle AB.BH
for AB is equal to BD ; and FH is the
square of AH ; therefore the rectangle
AB:BH is equal to the square of AH : ^
Wherefore the straight line AB is divided in H, so that the rectangle
AB.BH is equal to the square of AH; Which was to be done.

ELEMENTS

PROP. Xn. THEOR.

In obtuse angled triangles ^ if a perpendicular be drawn from any of
the acute angles to the opposite side produced^ the square of the side
subtending the obtuse angle is greater than the squares cf the sides
containing the obtuse angU^ by twice therectangle contained by the side
upon whichi when produced^ the perpendicular falls, and the straight
line intercepted between the perpendicular and the <^use angle.

Let ABC be an obtuse angled triangle, having the obtuse angle ACB,
and from the point A let AD be drawn (l2.^1.) perpendicular to EC
produced : The square of AB is greater than the squares of AC,
CB, by twice the rectangle BC.CD.

Because the straight iine BD is divided into two parts in the point
C, BD» = (4. 2.) BC« + CD« +
2BC.CD ; add AD' to both : Then
BD«4-AD« ?= BC«+eD« + AD«
+2BC.CD. But AB«=BD»+
AJy (47. 1), and AC»=€D'+
AD' (47. 1.) ; therefore, AB«=
BC'+AC* + 2BC.CD ; that is,
AB« is greater than BC«+AC»
by 2BC.CD. Therefore, in ob-
tuse angled triangles, dec. Q. E.D. S'

PROB. XIII. THEOR.

In every triangle the square of the side subtending any of the acute an-
glesy is less than the squares of the sides containing that angle, by
twice the rectangle contained by either of these sides, and the straight
line intercepted between the perpendicular, let fall upon it from the
opposite angle, and the acute angle.

Let ABC be any triangle, and the angle at B one of its acute aa-
gle^, and. upon BC, one of the sides containing it, let fall the per-
pendicular (12. 1.) AD from the opposite angle : The square of AC^
opposite to the angle B, is less than the squares of C6, BA by twice
the rectangle CB.BD.

First, let AD Ml within the trian-
gle ABC ; and because the straight
line CB is divided into two parts in
the point D (7. 2.), BC» + BD' =
2BC.BD+CD'. Add to each AD' ;
then BC + BD + AD'* = 2BC.BD
+CD' + AD'. But BD' + AD'
= AB', and CD' + DA' = AC
,(47. 1.) ; therefore BC= + AB' =
2BC.BD+AC'; that is, AC is less
than BC^+AB» by 8BC.BD.

. *

OP GEOMETRY. BOOK II.

Secondly, Let AD fall without the triangle ABC i*' Then because
the angle at D is a right angle, the angle ACB is greater (16. 1.) than
a riglit atigle, and AB«=(12. 2.) AC*+BC'+2BC.CD. Add BC
to each ; then AB'+BC* = AC^+2BC»+2BC.CD. But because
BD is divided into two parts in C, BC»+BC.CD=(3. 2.) BC.BD,
and 2BC* + 2BC.CD = 2BC.BD : therefore AB« + BC» = AC> +
2BC.BD ; and AC" is less than AB^+BC^ by 2BD.BC.

Lastly, Let the side AC be perpen-
dicular to BC ; then is BC the straight J^
line between the perpendicular and the
acute angle at B ; and it is manifest that
(47. L) AB'+BC»=AC»+2BC»=AC»
+2BC.BC. Therefore, in every trian-
gle^ dec. d. £. D.

PROP. XIV. PROB.

To describe a $guare that shall he eqmL to, a given rect&ineal figure.

Let A be 4he given rectilineal figure ; it is required to describe a
square that shall be equal to A.

Describe (45. L) the rectangular parallelogram BCDE equal to the
rectilineal figure A. If then the sides of it, BE, ED are equal to one
another, it is a square, and what was required is done ; but if they are
not equal, produce one of them, BE to F, and make EF equal to ED,
and bisect BF in 6 ; and from the centre 6, at the distance GB, or GF,
describe the semicircle BHF, and produce DE to H, and join GH.
Therefore, because the straight line BF is divided into two equal
parts in the point G, and into two unequal in the point £, the rect-
angle BE.EF, together with the square of EG, is equal (5. 2.) to
the square of GF : but GF is equal to GH ; therefore the rectangle
BE, EF, together with the square of EG, is equal to the square of
GH : But the squares of HE and EG are equal (47. 1.) to the square
of GH : Therefore also
' the rectangle BE.EF, to-
gether with the square of
EG, is equal to the squares
of HE and EG. Take
away the square of EG,
which is common to both,
and the remaining rect-
angle tflE.EF is equal to
the square of EH : But
BD is the rectangle con-

^ Sm 6gare of the lasTPropotitioar

ELEMENTS

tained by BE and EF, because EF is equal to ED ; therefore 6D ia
equal to the square of EH ; and BD is also equal to the rectilineal
figure A ; therefore the rectilineal figure A is equal to the square of
EH : Wherefore a square has been made equal to the given rectili-
neal figure A^ viz. the square described upon EH. Which was to be
done.

PROP. A. THEOR.

If one side efa triangle be bisec^y the sum of the squares of the other
two sides is double of the square ofhtdfthe side bisected^- and of the
square of the line draumfrom the point of bisection to the opposite aw
gleqfwe triangle.

Let ABC be a triangle, of which the side BC is bisected in D, and
DA drawn to the opposite angle ; The squares of BA aud AC are
together double of the squares of BD and DA.

From A draw AE perpendicular to BC, and because BE A is a right
angle, AB»=x(47. 1.) BE'+AE^and
AC»=CE«+AE^ wherefore AB'+
AC»=BE»+CE»+2AE«. But be-
cause the line BC is cut equally in D,
and unequally in E, BB»+CE*=
(9. 2,) 2BD»+2DE^ therefore AB'*
+AC»=3BD»+2DE^2AE«.

Now DE»+AE^=(47. 1.) AD',
and 2DE'+2AE»=2AD'; where-
foreJAB^ + AC» = 2BD« + 2AD^
Therefore, &c. Q. E. D.

PROP. B. THEOR.

The sum of the squares of the diameters of any parallelogram is equal to
the sum of the squares of the sides of the parallelogram.

Let ABCD be a parallelogram, of which the diameters are AC and
BD ; the sum of the squares of AC and BD is equal to the sum of
the squares of AB, BC, CD, DA. ,

Let AC and BD intersect one another in E : and because the ver-
tical angles AED, CEB are equal (15. 1.), and also the alternate an-
gles EAD, ECB (29. L), the triangles ADE, CEB have two angles in
the one equal iq two angles in the other, each to each ; but the sides
AD and BC, which are oppo.
site to equal angles in these
triangles, are also equal (34.
1.); therefore the other sides
which are opposite to the
equal angles are also equal
(26. L), viz. AE to EC, and
ED to EB.

OF GEOMETRY. BOOK II. 65

Since, therefore, BD is bisected in E, AB'+ AD'=(A. 2.) 2BE»
+2AE» ; and for the same reason, CD«+BC»=2BE2+2EC»=2BE*
+2AE>, because EC=AE. Therefore AB"+AD^+DCHBC2=
4BE«+4AE". But 4BE^=BD«, and 4AE2=:AC^ (2. Cor. 8. 2.) be-
cause BD and AC are both bisected in E ; therefore AB'+AC^^h
CD»+BC» = BD^+AC^ Therefore the sum of the squares, &c.
Q. E. D.

Cor. From this demonstration, it is manifest that the diameters of
tvery parall.elogram bisect one another.

ELEMENTS

GEOMETRY.

BOOK m.

DEFINITIONS.

The radius of a circle is the straight line drawn fr<Hn the cenir&
to the circumference. «

' I.

A straight line is said to touch
a circle^ when it meets the
circle, and being produced
does not cut it.

II.

Circles are said to touch one
another, which meet, but do
not cut one another.

III.

Straight lines are said to be equally dis-
tant from the centre of a circle, when
the perpendiculars drawn to them
from the centre are equal.

IV.

And the straight line on whieh'the great-
er perpendicular falls, is said to be
• . farther from the centre.

Aaarch of a circle is any part of the circumference,

A segment of a circle is the figure con-
tained by a straight line, and the arch
which it cuts off.

ELEMENTS OF GEOMETRY. BOOK HL 0*3:

An angle in a segment is the angle contain-
ed by two straight lines drawn from any
point in the circumference of the segment,
to the extremiti es of the straight line which
19 the basQ of the segment,

VII.

And an angle is said to insist or stand upon
the arch intercepted between the straight
lines which contain the angle.

VIII,

The sector of a circle is the ^figure con-
tained by two straight lines drawn from
the centre, and the arch of the circumfe-
rence between them.

IX.

Similar segments of a circle,
are those in which the an-
gles are equal, or which
contain equal angles. ^

PROP. I. PROB.

To find the centre of a given circle.

Let ABC be the given circle ; it is required to find its centre*
Draw within it any straight line.AB, and bisect (10. 1.) it in D ;
from the point D draw (11. 1.) DC at right angles to AB, and pro-
duce it to E, and bisect CE in F : the point F is the centre of the
circle ABC. .

For, if it be not, let, if possible, G be the centre, and jom IxA,
GD, GB : Then, because DA is equal to DB, and DG common to
the two triangles ADG, BDG, the two sides AD, DG are^qual to the
two BD, DG, each to each ; and the base ^

GA is equal to the base GB, because they
are radii of the same circle : therefore the
angle ADG is equal (8. 1.) to the angle
GDB : But when a straight line standing up-
on another straight line makes the adjacent
angles equal to one another, each of the an-
gles is a right angle (7. def. 1.) Therefore
tbe angle GDB is a right angle : But FDB is
likewise a right angle : wherefore the angle
FDB is equal to the angle GDB, the greater

68 ELEMENTS

to the less, which is impossible : Therefore G is not the centre oftte
circle ABC : In the same manner, it caa be shown that no other
point but F is the centre : that is, F is the centre of the circle ABC t
Which was to be found.

Cor. From this it is manifest that if in a circle a straight line bi-
sect another at right angles, the centre of the circle is in the line
which bisects the other.

PROP. II. THEOR.

I£any two points be taken in the circumference of a circle^ the straight
line which joins them shall fall within the circle.

Let ABC be a circle, and A, B any two points in the circumfe-
rence ; the straight line drawn from A to B r*
shall fall within the circle.

Take any point in AB as E ; find D the
centre of the circle ABC ; join AD, DB and
DE, and let DE) meet the circumference in
F. Then, because DA is equal to DB, tbe
angle DAB is equal (5. 1.) to the angle

DBA ; and because AE, a side of the trian- a/ NKTv/

gle DAE, is produced to B, the angle DEB -^^^C \ \^R
is greater (16. 1.) than the angle DAE ; but
DAE is equal to the angle DBE ; therefore
the angle DEB is greater than the angle DBE : Now to the greater
angle the greater side is opposite (19. 1.); DB is therefore greater
than DE : but BD is equal to DF ; wherefore DF is greater than
DE, and the point E is therefore within the circle. The same may
be demonstrated of any other point between A and B, therefore AB
is within the circle. Wherefore, if any two points, <&c. Q. E. D»

PROP. III. THEOR.

If a straight line drawn through the centre oj a circle bisect a straight
line in the circle^ which does not pass through the centre^ it will cut
that line at right angles ;, and ifitculitat right angles, it wUl 6t-
sect it.

Let ABC be a circle, and let CD, a straight line drawn through the
ceptre, bisect any straight line AB, which does not pass through the
centre, in the point F ; It cuts it also at right angles.

Take (L 3.) E the centre of the circle, and join EA, EB. Then
because AFis equal to FB,and FE common to the two triangles AFE,
BF£, there are two sides in the one equal to two sides in the other ;

OF GEOMETRY. BOOK III.

6§

but the base EA is equal to the base £B ;
therefore the angle AFE is equal (8. l.)to
the angle BFE« And when a straight line
standing upon another makes the adjacent
angles equal to one another, each of them
is a right (7. Def. 1.) angle : Therefore
each of the angles AFE, BFE is a right
angle ; wherefore the straight line CD, » ^
dr^wn through the centre bisecting AB, -^
which does not pass through the centre, ^
cuts AB at right angles.

Again, let CD cut AB at right angles ; CD also bisects AB, that
is, AF is equal to FB.

The same construction being made, because the radii EA, EB are
equal to one another, the angle EAF is eoual (§; 1.) to the angle £BF ;
and the right angle AFE is equal to the right angle BFE : Therefore,
in the two triangles EAF, EBF, there are two angles in one equal
to (wo angles in the other ; now the side EF, which is opposite to
one of the equal angles in each, is common to both ; therefore the^
other sides are equal to (26. 1.) : AF therefore is equal to FB.
Wherefore, if a straight line, &c. Q. Ev"D.

PROP. IV. THEOR.

If in a circle two straight lines cut one another^ which do not both pass
through the centre, they do not bisect each other.

Let ABCD be a circle, and AC, BD two straight lines in it, which
cut one another in the point E, and do not both pass through the cen-
tre : AC, BD do not bisect one another.

For if it is possible, let AE be equal to EC ,and BE to ED ; If one
of the lines pass through the centre, it is ■*
plain thatit cannot be bisected by the other,
which does not pass through the centre. But
if neither of them pass through the centre, A^
take (1. 3.) F the centre of the circle, and
join EF : and because FE, a straight line
through the centre, bisects another AC,
which does not passthrough the centre, it
must cut it at right (3. 3.) angles ; where-
fore FEA is a right angle. Again, because
the straight line FE bisects the straight line BD, which does not pass
through the centre, it must cut it at right (3. 3.) angles ; wherefore
FEB is a right angle : and FEA was shown to be a right angle :
therefore FEA is equal to the angle FEB, the less to the greater,
which is impossible ; therefore AC, BD, do not bisect one another.-
Wherefore, if in a circle, &c. Q. E. D. ,

70 ELEMENTS^

PROP. V. THEOR.

If ttoo circles cut one ttnother, they cannot have the same centre:

Let the two'circles ABC, CD6 cut one another in the point B, C ;:
they have not the same centre.

For, if it be possible, let E be their centre : join EC, and draw any
straight line EFG meeting the circles
in F and G : and because E is the cen-
tre of the circle ABC, CE is equal to
EF : Again, because E is the centre
of the circle CDG, CE is equal to
EG : but CE was shown to be equal
to EF, therefore EF is equal to EG,
the less to the greater, which is im-
possible : therefore £ is not the
centre of the circles ABC, CDG.
Wherefore, if two circles,&c. Q.E.D.

PROP. VI. THEOR.

Iftujo circles touch one another internally y they cannot have the same

centre*

Let the two circles ABC, CDE, touch one another internally in the
point C ; they have not the same centre.

For, if they have, let it be F ; join FC, and draw any straight line
FEB meeting the circles in E and B ; and q ,

because F is the centre of the circle ABC,
CF is equal to FB ; also, because F is the
centre of the circle CDE, CF is equal to
FE : but CF was shown to be equal to FB ;
therefore FE is equal to FB, the less to the

greater, which is impossible ; Wherefore F \ \ ^^^^r^

is not the centre of the circles ABC, CDE. J^ ^ ^ '^

Therefore, if two circles, <fec. Q. E. D.

PROP. VIL THEOR.

If any point be taken in the diameter of a circle which is not the centre of
all the straight lines which can be drawn from it to the circumference^
the greatest is that in which the centre is, and the other part of that di^
ameter is the least ; and, of any others, that which is nearer to the line
passing through the centre is always greater than one more remote from
it ; And from the same point there can be drawn only two straight lin€9
that are equal to one another, one upon each side of the shortest line.

Let ABCD be a circle, and AD its diameter, in which let any point
F be taken which ie not the centre : let the centre be E ; of aU the

OP GEOMETRY. BOOK III.

islraight lines FB^ FC, FG, <Sz;c. that can be drawn from F to the ctr*
Gumference, FA is the greatest ; and FD, the other part of the dia-
meter ADf is the least ; and of the others, FB is greater than FC,
and FC than FG.

Join BE, CE, GE ; and because two sides of a triangle are greater
(20. 1.) than the third, BE, EF are greater than BF ; but AE is equal
to EB ; therefore AE and EF, that is, AF, is greater than BF : again,
because BE is equal to CE, and FE common to the triangles BEF,
CEF, the two sides BE, EF are equal to
the twi) CE, EF : but the angle BEF is
greater than the angle CEF ; therefore the
base BF is greater (24. 1.) than the base
FC ; for the same reason, CF is greater
than GF. Again, because GF, FE are
greater (20. 1.) than EG, and EG is equal
to ED ; GF, FE are greater than ED ; take
away the common part FE, and the re-
mainder GF is greater than the remainder
FD : therefore FA is the greatest, and FD
the least of all the straight lines from F to the circumference ; aiid
BF is greater than CF, and CF than GF.

Also there can be drawn only two ec[ual straight lines from the
point F to the circumference, one upon each side of the shortest line
FD : at the point E in the straight hne EF, make (23. 1.) the angle
FEH equal to the angle GEF, and join FH : Then, because GE is
equal to EH, and EF common to the two triangles GEF, HEF ; the
two sides GE, EF are equal to the two HE, EF ; and the angle GEF
is equal to the angle HEF ; therefore the base FG is equal (4. 1.)
to the base FH : but besides FH, no straight line can be drawn from
F to the circumference equal to FG : for, if there can, let it be FK ;
and because FK is equal to FG, and FG to FH, FK is equal to FH ;
that is, a line nearer to that which passes through the centre, is equal
to one more remote, which is impossible. Therefore, if any point be
4aken, &;c. Q. E. D.

PROP. Vni. THEOR.

If any point he taken toithout a circle, and straight lines he drawn from
it to the circumference^ whereof one passes through the centre ; of those
which faU upon the concave circumference, the greatest is thai which
passes through the centre; and of the rest that which is nearer to that
through the centre is always greater than the more remote; But of
those which faU upon the convex circumference, the least is that he''
tween the point toithout the circle, and the diameter ; and of the rest,
that which is nearer to the least is always less than the more remote :
And only two equal straight lines can he drawn from the point unto
the circumference^ one upon each side of the least.

Let ABC be a circle, and D any point without it, from which let the

£L£:ments

atraight lines DA, DE, DF, DC be drawn to the circumferenc^i
whereof DA passes through the centre. Of those which fall upon
the concave part of the circumference AEFC, the greatest is AD,
which passes through the centre ; and the line nearer to AD is always
greater than the more remote, viz. DE than DF, and DF than DC ;
but of those which fall upon the convex circumference HLKG, the
least is D6, between the point D and the diameter AG ; and the near-
er to it is always less than the more remote, viz. DK than DL, and
DL than DH.

Take (1. 3.) M the centre of the circle ABC, and join ME, MF,
MC, M K, ML, MH : And because AM is equal to ME, if MD be add*
ed to each, AD is equal to EM and MD ; but EM and MD are greater
(20. 1 .) than ED ; therefore also AD is greater than ED. Again, be*
cause ME is equal to MF, and MD common to the triangles EMD,
FMD ; EM, MD are equal to FM, MD ; but the angle EMD is great-
er thad the angle FMD ; therefore
the base ED is greater (24. 1.) than
the base FD. In like manner it may
be shewn that FD is greater than
CD. Theref<»^e DA is the greatest ;
and DE greater than DF; and DF
than DC.

And because MK, KD are greater
(20. 1.) than MD, and MK is equal
to MG, the remainder KD is greater
(5. Ax.) than the remainder GD, that
is, GD is less than ED: And because
MK, DK are drawn to the p^int K
within the triangle MLD from M,
D, the extremities of its side MD ;
MK, KD are less (21. 1.) than ML,
LD, whereof MK is equal to ML ;
therefore the remainder DK is less
than the remainder DL : In like
manner, it may be shewn that DL is less than DH : Therefore DG
is the least, and DK less than DL, and DL than DH. '

Also there can be drawn only two equal straight lines from the
point D to the circumference, one upon each side of the least : at
the point M, in the straight line MD, make the angle DMB equals to
the angle DMK, and join DB ; and because in the triangles KMD,
BMD, the side KM is equal to the side BM, and MD common to both,
and also the angle KMD equal to the angle BMD, the base DK is
equal (4. 1.) to the base DB. But, besides DB, no straight line can
be drawn from D to the circumference, equal to DK : for, if there
can, let it be DN ; then, because DN is equal to DK, and DK equal
to DB, DB is equal to DN ; that is, the line nearer to DG, the least,
equal to the more remote, which has been shewn to be impossible.
If, therefore, any point, &c. - ' Q. E. D.

OF GEOMETRY. BOOK III.

PROP. IX. THEOR.

tf a point be taken within a ctrdeyfromithich there faU more than two
equal straight lines upon the drcuniferencey that point is the centre of
the circle.

Let the point D be taken within the circle ABC, from which there
fall on the circumference more than two equal straight lines, viz. DA,
DB, DC, the point D is the centre of the circle. .

For, if not, let E be the centre, join DE, and produce it to the cir-
cumference in F,' G ; then FG id a diame-
ter of the circle ABC : And because in
FG, the diameter of the circje ABC, there
is taken the point D which is not the cen-
tre, E>G is the greatest line from it to the
circumference, and DC gre^iter (7. 3.)
than DB^ and DB than DA ; but they
■ are likewise equal, which is impossible :
Therefore E is not the centre of the cir-
cle ABC : In like manner it may be de-
monstrated, that no other point but D is
the centre : D therefore is the centre.
IVherefore,. if a point betaken, (be. Q. E. D.

PROP. X. THEOR.

One circle cannot cut another in jnore ikan /tc^ points.

If it be possible, let the circumference FAB cut the circumference
DEF in more than two points, viz. in B, G, F ; take the centre K of
the circle ABC, and join KB, KG, KF ; and because within the circle
DEF there is taken the poinl K, from which more than two equal
straight lines, viz. KB, KG, KF, fall on
tha circumference DEF, the point K is
(9. 3.) the centre of the circle DEF ;
but K is also the centre of the circle
ABC ; therefore the same point is the
centre of two circles that cut one ano-
ther, which is impossible (5. 3.). There-
fore one circumference of a circle can-
not cut another in more than two points.
Q. E. D.

PROP. XL THEOR.

«
Iftioo circles touch each other inlemally, the straight line which joins
their centres Jteing produced, will pass through the point of contact.

Let the two circles ABC, ADE, touch each other internjilly in the

ELEMENTS

point A, and let F be the centre of the circle ABC, and G the centre
of the circle ADE ; the straight line
which joins the centres F, G, being pro*
duced, passes through the point A.

For, if not, let it fall otherwise, if
possible, as FGDH, and join AF, AG :
And because AG, GF are greater (20. 1.)
than FA, that is, than FH, for FA is equal
to FH, being radii of the same circle ;
take away the common part FG, and the
remainder AG is greater than the remain-
der GH. But AG is equal to GD, there-
fore GD is greater than GH ; and it is
also less, which is impossible. Therefore the straight line which joinii
the points F and G cannot fall otherwise than on the point A ; that is,
it must pass through A. Therefore, if two circles, &c. Q. £. D.

PROP. XIL THEOR.

If two circles touch each ether eaftermdly, the straight line which joins
their centres wiU pass through the point ^ contact*

Let the two circles ABC, ADE, touch each«other externally in the
point A ; and let F be the centre of the circle ABC, and G the centre
of ADE ; The straight line which joins the points F, G shall pass
through the point of contact.

For, if not, let it pass otherwise, if possible, FCDG, and join FA,
AG : and because F is the centre of the circle ABC, AF is equal to
FC : Also because G
is the centre of the
circle, ADE, AG is
equal to GD. There-
fore FA, AG are .equal
to FC, DG; where-
fore the whole FG is
greater than FA, AG ;
but it is also less (20.
1.), which is impossi-
ble : Therefore the
straight line which joins

the points F, G cannot pass otherwise than through the point of con-
tact A ; that is, it passes through A. Therefore, if two circles, &g*
Q. E. D.

PROP. XIIL THEOR.

One circle cannot touch another in more points than one, whether U

touches it on the inside or outside.

For, if it be possible, let the circle EBF touch the circle ABC ift

OF GEOMETRY. BOOK III.

fiKn'e points than one, and first on the inside, in the points B, D ; join
BD, and draw (10. 11. 1.) GH, bisecting BD at right angles : There-
fore because the points B, D are in the circumference of each ofthe

circles, the straight line BD falls within each (2. 3.) of them : and
therefore their centres are (Cor. 1. 3.) in the straight line GH which
bisects BD at right angles : Therefore GH passes through the point
of contact ( 1 1 . 3.^ ; but it does not pass through it, because the points
B, P are without the straight line GH, which is absurd : Therefore
one circle cannot touch another in the inside in more points than one.
Nor can two circlets touch ojie another on the outside in more than
one point : For, if it be possible, let the circle ACK touch the circle
ABC id the points A, C, and join AC :
Therefore, because the two points A, C are
in the circumference of the circle ACK,
the straight line AC which joins them
shall fall within the circle ACK : And
the circle ACK is without the circle ABC ;
and therefore the straight line AC is also
without ABC ; but, because the points
A> C are in the circumference of the cir-
cle ABC, the straight line AC must be
within (2. 3.) the same circle, which is
absurd : Therefore a circle cannot touch
another on the outside in more than one
point : and it has been shewn, that a cir-
cle cannot < touch another on the inside in
more than one point. Therefore, one circle, dec. Q. E. D.

PROP. XIV. THteOR.

Equal straight lines in a circle are equally distant from tJie centre;
and those which are equally distant from the centre^ are equal to one
another*

Let the straight lines AB, CD, in the circle ABDC, be equal to one
another : they are equally distant from the centre.

ELEMENTS

Take £ the ceDtre or the circle ABDC, and from it draw EF, EG,
perpendiculars to AB, CD ; join AE and EC. Then, because the
straight line EF passing through the cen-
tre, cuts the straight line AB, which do^s
not pass through the centre at right an.
gles, it also bisects (3. 3.) it : Where* Aj
£>re AF is equal to FB, and AB double
of AF. For the same reason, CD is
double of CG : But AB is equal to CD ;
therefore AF is equal to CG : And be-
cause AE is equal to EC, the square of
AE is equal to the square of EC : Now
the squares of AF, F£..are equal (47. 1.)
to the square of AE, because the angle AFE is a right angle ; and,
lor the like reason, the squares of EG, GC are equal to the square
of EC : Therefore the squares of AF, FE are equal to the squares of
CG, GE, of which the square of A F is equal to the square ef CG, be*
cause AF is -equal to CG ; therefore the remaining square of FE is
equal to the remaining square of £1G, and the straight line £F is there-
fore equal to EG :. But straight lines in a circle are said to be equally
distant from the centre when the perpendiculars drawn to them front
the centre are equal (3. Def. 3.) i therefore AB, CD are equally dis-
tant from the centre.

Next, if the straight lines AB, CD be equally distant from the
centre, that is, if FE be equal to EG, AB is equal to CD. For, the
same construction being made^ it may,, as before, be demonstrated!,
that AB is double of AF, and CD double of CG, and that the squares
of EF, FA are equal to the squares of EG, GC ; of which the square
of FE is equal to the square of EG,, because FE is equal to EG* ;
therefore the remaining square of A F is equal to the remaining square
of CG ; and the straight line AF is therefore equal to CG : But AH
is double of AF, and CD double of CG ; wherefore AB is equal to
CD. Thereibre equal straight lines, dz^c. Q. E.. D.

PROP. XV. THEOR.

The diameter is the greatest itraigJu Une in a circle ; and of aittotherf,
that which is nearer to the centre is always greater than one more re*-
mole ; and the greater is nearer to the centre than the less.

Let ABCD be a circle, oC which the di-
ameter is AD, and the centre E ; and let
BC be nearer to the centre than FG ; AD '
is greater than any straight line BC which -^j
is not a diameter, and BC greater than FG.

From the centre draw EH, EK perpen-
diculara to BC, FG, and join EB, EC, EF ;
and because AE is equal to EB, and ED to
EC, AD is equal to EB, EC : But EB, EC
are greater (20. 1.) than BC ; wherefore,
«lsO| AD is greater thaa BC

OP GEOMETRY, BOOK HI.

And, ^because BC is nearer to the centre than FG, EH is less
(4. Def. 3.^ than EK ; But, as was demonstrated in the precediifg,
BC is double of^BH, and FG double of FK, and the squares of EH,
HE are equal to the squares of EK, KF, of which the square of EH
is less than the square of EK, because EH is less than EK ; there-
fore the square of BH is ^greater than the square of FK, and the
straight line BH greater than FK ; and therefore BC is greater than
FG. '

Next, let BC be greater than FG ; BC is nearer to the centre than
FG : that is, the same construction being made, EH is less than EK ;
Because BC is greater than FG, BH likewise is greater than KF :.
but the squares of BH, HE are equal to the squares of FK, KE, of
which the square of BH is greater than the square of FK, because
BH is greater than FK ; therefore the square of EH is less than the
square of EK, and the straight line EH less than EK. Wherefores
the diameter, dec. Q. E. D.

PROP. XVI. THEOR.

The straigU line drawn at right angles to the diameter of a circle, from
the extremity of it, falls without the circle ; and no straight line cam
he dravm between that straight line and the cvrcumference, from the
extremity of the diameter, so as not to cut the circle.

Let ABC be a circle, the centre of which is D, and the diameter
AB : and let AE be drawn from A perpendicular to AB, AE shall
fall without the circle.

In AE take any point F, join DF and let DF meet the circle in C.
Because DAF is a right angle, it is
greater than the angle AFD (32. 1.) ; '
but the greater angle of any triangle
is subtended by the greater side (19.
1.), therefore DF is greater than DA ;
now DA is equal to DC, therefore
DF is greater than DC, and the point
F is therefore without the circle.
And F is any point whatever in the
line AE, therefore AE falls without
the circle.

Again, between the straight line
AE.and the circumference, no straight
line can be drawn from the point A,
which does not cut the circle. Let AQ be drawn in jhe angle DAE :

ELEMENTS

from D draw DH at right angles to AG ;
and because the angle DHA is a right
angloy and the ancle DAH less than a
rignt angle, the side DH oT the triangle
DAH is less than the side DA (19. 1.).
The point H, thereforOf is within the
circle, and therefore the straight line AG
cuts the circle. B

Cob. From this it is manifest, that the
straight line which is drawn at right an-
gles to the diameter of a circle from the
extremity of it, touches the circle ; and
that it touches it only in one point ; because, if it did meet the circle
in twot it would fall within it (2. 3.). Also it is evident that there can
be but one straight line which touches the circle in the same point.

PROP. XVII. PROB.

To draw a straight line from a given point either tnthout or in the cir^
cumference^ which shall touch a given circle.

First, Let A be a given point without the given circle BCD ; it is
arequired to draw a straight line from A which shall touch the circle.

Find (1* 3*) the centre E of the circle, and join AE ; and from the
centre E, at the* distance EA, describe the circle AFG ; from the
point D draw (11.1.) DF at right angles to EA, join EBF, and draw
A^. AB touches the circle BCD. *

Because E is the ceatre of the cir-
cles BCD, AFG, EA is equal to EF,
and ED to EB ; therefore the two
sides AE, EB are equal to the two FE,
ED, and they contain the angle at E
common to the two triangles AE6,
FED; therefore the base DFis equal
to the base AB, and the triangle FED
to the triangle AEB, and the othet* an-
gles to the other angles (4. 1.) ; There-^
fore the angle EBA is equal to the an-
gle EDF; but EDF is a right angle, wherefore EBA is a right an-
gle ; and EB is a line drawn from the centre : but a straight line
drawn from the extremity of a diameter, at right angles to it, touches
the circle (Cor. 16. 3.) : Therefore AB touches the circle ; and is
drawn from the given point A. Which was to be done.

But if the given point be in the circumference of the circle, as the
point D, draw DE to the centre £, and DF at right angles to DE ;
jbP touches the circle (Cor. 16. 3.)

OP GEOMETRY. BOOK in.

PROP. XVm. THEOR.

If a straight line touch a circle, the straight line drStmfrom the centre
to the point of contact j is perpendicular to the Une touching the circle.

Let the straight Jine DE touch the circle ABC in the point C ; take
the centre F, and draw the straight Une FC : FC is pefpendicular
to DE.

For, if it be not, from the point F draw FBG perpendieulartb DE>;
and because FGC is a right angle, GCF n^ust be (17. 1.) an acute «n>
gle ; and to the greater angle the great- a^

er (19. 1.) side is opposite : Therefore
FC is greater than FG ; but FC is equal
to FB ; therefore FB is greater thai;i FG,
the less than the greater, which is im-
possible ; wherefore FG is not perpen.
dicular to DE : In th^ same mtoner it
may be shewn, that no other line but
FC can be perpendicular to DE ; FC is
therefore perpendicular to DE. There-
fore, if a straight line, &c. Q. E% D. j)

PROP. XIX. THEOR.

If a straight line touch a circlCy and from the point of contact a straight
line be drawn at right angles to the touching line, the centre of the
circle is^ in that line.

Let the straight line DE touch the circle ABC, in C, and from O
let CA be drawn at right angles to DE ; the centre of the circle is in
CA.

For, if not, let F be the centre, if possible, and join CF. Bepaase
DE touches the circle ABC, and FC is
drawn from the centre to the point of
contact, FC is perpendicular (13; 3.) to
DE ; therefore FCE is a right angle ;
But ACE is also a right angle ; therefore
the angle FCE is equal to the angle
ACE, the less to the greater, which is im- ;g|
possible ; Wherefore F is not the centre
of the circle ABC : In the same manner
it may be shewn, that no other point
which is not in CA, is the centre ; that is, ^r
the centre is in CA. Therefore, if a
straight line, &c. Q. E. D.

ELEMENTS

PROP. XX. THEOR.

T*he angle at (he centre of a circle is dauMe of the angle at the ctrcum^
ference, upon the same base, that is, ispon the same part of tht circum'
ferenee.

. Let ABC be a circloy and BDC an angle at the centre, and BAG
an angle at the circumference which have the sMne circumference
BC for the base ; the angle BDC is double of the angle BAG.

First, let D, the centre of the circle, be within the angle BAG, and
join AD, and produce it to E : Because DA is equal to DB, the angle
DAB is equal (5. 1.) to the angle DBA :
therefore the angles DAB, DBA together
are double of the angle DAB; but the an-

fie BDE is equal (32. 1.) to the angles DAB,
>BA; therefore also the angle BDE is
double of the angle DAB ; For the same
reason, the angle EDG is double of the
angle DAG : Therefore the whole angle
BDC is double of the whole angle BAG.

Again, let D, the cetitre of the circle,
be without the angle BAG ; and join AD
and produce it to E. It may be demon-
strated, as in the first case, that the an-
gle EDG is double of the Angle DAG,
and that EDB, a part of the first, is dou-
ble of DAB, a part of the other ; there-
fore the remaining angle BDG is double
of the remaining jingle BAG. Therefore
the angle ^t the centre, ^c. Q. E. D.

PROP. XXI. THEOR.

: The angles in the same segment of a circle are equal to one another.

Let ABGD be a circle, and BAD, BED angles in the same seg-
ment BAED : The angles BAD, BED are ^ _j;
equal to one another.

Take F the centre of the circle ABGD :
And, first, let the segment BAED be great-
er than a semicircle, and join BF, FD :
And because the angle BFD is at the cen-
tre, and the angle BAD at the circumfer-
ence, both having the same part of the -^
circumference, viz. BCD, for their base ;
therefore the angle BFD is double (20.
3.) of the angle BAD : for the same rea- C

- \

OF GEOMETRY. BOOKvIIL 81

son, the angle BFD is -double of the angle BED : Therefore the an-
gle BAD is equal to the an^le BED.

But, if the segment BAED be not greater
than a semicircle, let BAD, BED be an-
^gles in it ; these also are equal to one ano-
ther. Draw AF to the centre, and produce B/
to C, and join Cg : Therefore the seg-
ment BADC is greater than a semicircle ;
and the angles in it, BAG, BEC are equal,
by the first case : For the same reason,
^ because CBED is greater than a semicir-
cle, the angles CAD, CED are equal;
Therefore the whole angle BAD is equal
to the whole angle BED. Wherefore the angles in the same seg-
ment, dec. Q. £. D.

PROP. XXII. THEOR.

The opposite angles of any quadrilateral figure described in a circle^

are together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle A BCD ; any two
of its opposite angles are together equal to two right angles.

Join AC, BD. The angle CAB is equal (21. 3.) to the angle
CDB, because they are in the same seg-
ment BADC, and the angle ACB is equal
to the angle ADB, because they are in
the< same segment ADCB ; therefore the
whole angle ADC is equal to the angles
CAB, ACB : To each of these equals
add the angle ABC ; and the angles ABC,
ADC, are equal to the angles ABC, CAB,
BCA. But ABC> CAB, BCA are equal
to two right angles (32. 1.) ; therefore
also the angles ABC, ADC are equal to
two right angles ; In the same manner, the angles BAD, DCB may
be shewn to be equal to two right angles. Therefore the opposite an-
gles, &c. Q. E. D.

PROP. XXIII. THEOR.

Upon tJie same straight line, and upon the same side of it, there cannot
he two similar segments of circles ^ not coinciding with one another.

If it be possible, let the two similar segments of circles, viz. ACB,
ADB, be upon the same side of the same straight line AB, not coin-
ciding with one another ; then, because the circles ACB, ADB, cut
one another in the two points A, B, they cannot cut one another in

82 ELEMENTS

any other point (10. 8.) : one of the seg-
ments must therefore fall within the other :
let ACB fall within ADB, draw the straight
line BCD, and join GA, DA : and because
the segment ACB is similar to the segment
ADB, and similar segments of circles con-
tain (9. def. 3.) equal angles, the angle ^
ACB is equal to the angle ADB, the exte-
rior to the interior, which is impossible (16. 1»). Therefore, the>i&
cannot be two similar segments of circles upon the same side of the
same line^ which do not coincide* Q. £. D.

PROP. XXIV. THEOR.

Similar segments- of circles upon equal straight lines are equaSL to one

anoi?ier.

Let AEB, CFD be similar segments of circles upon th^' equal
straight lines AB, CD; the segment AEB is equal to the segment CFD.

For, if the segment AEB be applied to the segment CFD, so as the
point A be on C, and iji -w^

the straight line A B ^ -^ '

upon. CD, the point
B shall coincide with
the point D, because
AB is equal to CD : A B C D

Therefore the straight line AB coinciding with CD, the segment AEB
must (23. 3.) coincide with the segment CFD, and therefore is equal
to it. Wherefore, similar segments, 6ic, Q. £. D.

PROP. XXV. PROB.

A segment of a circle being given, to describe the circle of which it is:

the segment.

Let ABC be the given segment of a circle ; it is required to de-
scribe the circle of which it is the segment.

Bisect (10. 1.) AC in D, and from the point Ddraw (11. 1.) DB at
right angles to AC, and join AB : First, let the angles ABD, BAD be
equal to one another ; then the straight line BDis equal (6. 1.) to DA,
and therefore to DC ; and because the three straight lines DA, DB,
DC, are all equal ; D is the centre of the circle (9. 3.) : from the
centre D, at the distance of any of the three DA, DB, DG, describe a
circle ; this shall pass through the other points ; and the circle of
which ABC is a segment is described : and because the centre D is
in AC, the segment ABC is a semicircle. Next, let the angles ABD,
BAD be unequal ; at the point ^^ in the straight line AB, make (23.
1 .) the angle BAE equal to ^he angle ABD, and produce BD, if neces-

OF GEOMETRY. BOOK III.

sary, to E, and join EC : and because the angle ABE is equal to the
angle BAE, the straight line BE is equal (6. 1.) to EA : and because
AD is equal to DC, and DE common to the triangles ADE, CDE, the
two sides AD, DE are equal to the two CD, DE, each to each ; and
the angle ADE is equal tolhe angle CDE, for each of them is a right
angle ; therefore the base AE is equal (4. 1.) to the base EC : but
AE was shewn to be equal to EB, wherefore also BE is equal to EC :
and the three straight lines AE, EB, EC are therefore equal to one
another ; wherefore (9. 3.) E is the centre of the circle! From the
centre E, at the distance of any of the three AE, EB, EC, describe
a circle, this shall pass through the other points ; and the circle of
which ABC is a segment is described : also, it is evident, that if the
angle ABD be greater than the angle BAD, the centre E falls without
the segment ABC, which therefore is less thaq a semicircle ; but if
the angle ABD be less than BAD, the centre E falls within the seg-
ment ABC, which is therefore greater thaaa semicircle : Wherefore,
a segment of a circle being given, the circle is described of which it
is a segment. Which was to be done.

PROP* XXVI. THEOR.

In equdt circles, equal angles stand upon equal arches, whether they be

at the centres or circumferences.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF
at their centres, and BAC, EDF at their circumferences : the arch
BKC is equal to the arch ELF.

Join BC, EF ; and because the circles ABC, DEF are equal, the
straight lines drawn from their centres are equal ; therefore the two

C E

k ■

M ELEMENTS

sides BG, GC, are equal to the two EH, HF ; and the angle at G is
equal to the angle at H ; therefore the ba^e BC is equal (4. 1.) to the
base EF :' and because the angle at A is equal to the angle at D, the
segment BAG is similar (9. def. 3.) to the segment EDF ; and they
are upon equal straight lines BC, EF ; but similar segments of circles
upon equal straight lines are equal (24. 3.) to one another, therefore
the segipent BAG is equal to the segment EDF : but the whole circle
ABC is equal to the whole DEF ; therefore the remaining segment
BKC is equal to the remaining segment ELF, and the arch BKC to
the arch ELF. Wherefore, in equal circles, &c. Q. E. D.

PROP. XXVIL THEOR.

Ill equal cirdesy the angles whieh etand upon equal arches are equal to
one another y whether they he at the centres or circumferences*

Let the angles BGC, EHF at the centres, and BAG, EDF at the
circumferences of the equal circles ABC, D£F stand upon the equal
arches BC, EF: the angle BGC is equal to the angle EHF, and the
angle BAG to the angle EDF.

If the angle BGC be equal to the angle EHF, it is manifest (20. 3.)
that the angle BAG is also equaKto EDF. But, if not, one of them is
the greater : let BGC be the greater, and at the point G, in the straight
line BG, make the angle (23. 1.) BGK equal to the angle EHF. And
because equal angles stand upon equal arches (26. 3.), when they are
1^ at the centre, the arch BK is equal to the arch EF : but EF is equal

to BC ; therefore also BK is equal to BC, the less to the greater,
which is impossible. Therefore the angle BGC is not unequal to the
angle EUF \ that is, it is equal to it : and the angle at A is half the

angle BGC, and the angle at D half of the angle EHF ; therefore
the angle at A is equal to the angle at D. Wherefore, in equal cir-
cles, &c. Q. E. D.

PROP. XXVHL / THEOR.

In equal circles, equal straight lines cut off equal arches, the greater
equal to the greater, and the less to the less.

Let ABC, DEF be equal circles, and BC, EF equal straight lines in

OF GEOMETRY, BOOK IH.

them, which cut off the two greater arches BAC, EDF, and the two
less BGC, EHF : the greater BAC is equal to the greater £DF, and
the less BGC to the less EHF.

Take (1. 3.) K, L, the centres of the circles, and join BE, KG, EL,
LF ; and because the circles are equal, the straight lines from their

centres are equal; therefore BK, R€ are equal to EL, LF ; buf the
base BC is also equal to the base EF ; therefore the angle BKC is
equal (8.1.) to the angle ELF : and equal angles stand upon equal
(26. 3.) arches, when they are at the centres ; therefore the arch
BGC is equal to the arch EHF. But the whole circle ABC is equal
to the whole EDF ; the remaining part, therefore, of the circumfer-
ence, viz. BAC, is equal to the remaining part EDF. Therefore, in
equal circles, &c. Q. E. B.

PROP. XXIX. ^ THEOR.
In egtud ^cles eqwd arches' are subtended hy equal straight lines.

Let ABc/dEF be equal circles, and let the arches BGC, EHF
alsQ be equal ; and join BO, EF : the straight- line BC is equal to
the straight line EF.

Take (L 3.) K, L the centres of the circles, and join BK, KC, EL,
LF : and because the arch BGC is equal to the arch EHF, the angle
BKC is equal (27. 3.) to the angle ELF : also because the circles
ABC, DEF ar^ equal, their radii are equal : therefore BK, KC are

86 ELEMENTS

equal to EL, LF : and they contain equal angles ; therefore the base
BC is equal (4. 1.) to the base EF. Therefore, in equal circles, &c.
Q. E. D.

PROP. XXX- THEOR.
To bisect a given arch, that is, to divide it into ttoo equal parts.

Let ADB be the given arch ; it is required to bisect it.

Join AB, and bisect (10. 1.) ifin C ; from the point C draw CD
at right angles to AB, and join AD, DB : the arch ADB is bisected
in the point D.

Because AC is equal to CB, and CD common to the triangle ACD,
BCD, the two sides AG, CD are equal to* -r\

the two BC, CD ; and the angle ACD is -*-'

equal to the angle BCD, because each of
them is a right angle : therefore the base

AD is equal (4. L) to the base BD. But

equal straight lines cut off equal (28. 3.) J^

arches, the greater equal to the greater, ^

and the less to the less ; and AD, DB are each of them less than a

semicircle, because DC passes through the centre (Cor. 1.3.);

wherefore the arch AD is equal to the arch DB ; and therefore the

given arch ADB is bisected in D. Which was to be done.

PROP. XXXI. THEOR.

In a circle, the angle in a semicircle is a right angle ; hut the angle in a
segment greater than a semicircle is less than a right angle ; and the
angle in a segment less than a semicircle is greater than a right angle.

Let ABCD bo a qircle, of which the diameter is BC, and feentre
E ; draw CA dividing the circle into the segments ABC, ADC, and
join BA, AD, DC ; the angle in the semicircle BAC is a right angle ;
and the angle in the segment ABC, which is greater than a semicir-
cle, is less than a right angle ; and the angle in the segment ADC,
which is less than a semicircle, is greater than a right angle.

Join A£, and produce BA to F ; and because BE is equal to EA,
the angle EAB is equal (5. 1.) to EBA : -uk

.also, because AE is equal to EC, the

angle EAC is equal to ECA ; where- A.

fore the whole angle BAC is equal to
the two angles ABC, ACB. But FAC,
the exterior angle of the triangle ABC,
is also equal (32. 1.) to the two angles ^x \ ' Nj^

ABC, ACB; therefore the angle IB AC Bp ^^ ^C

is equal to the angle FAC, and each of
them is therefore a right (7. def. 1.)
angle ; wherefore the angle BAC in a
semicircle is a right angle.

OF GEOMETRY. BOOK III.

And because the two angles ABC, BAG of the triangle ABC are
together less (17. 1.) than two right angles, and BAG is a right angle,
ABC must be less than a right angle ; and therefore the angle in a
segment ABC, greater than a semicircb, is less than a right angle.

Also because ABCD is a quadrilateral figure in a circle, any two
of its opposite angles are equal (22. 3.) to two right angles ; there-
fore the angles ABC, ADC are equal to two right angles ; and ABC
is less than aright angle ; wherefore the other ADC is greater than
a right angle. Therefore, in a circle, &c. Q. E. D.

Cob. From this it is manifest, that if one angle of a triangle be
equal to the other two, it is a right angle, because the angle adjacent
to it is equal to the same two ; and when the adjacent angles are
equal, they are right angles.

PROP. XXXII. THEOR.

If a straight line touch cLcircle, and from the point of contact a straight
line he drawn cutting the circle^ the angles made by this line jcith the
line which touches the circle, shall be equal to the angles in the aUer»
note segments of the circle.

Let the straight line £F touch the cjrcle ABCD in B, and from the
point B let the straight line BD be drawn cutting the circle : The
angles which BD makes with the touching line EF shall be equal to
the angles in the alternate segments of the Circle : that is, the angle
FBD is equal to the angle which is in the segment DAB, and the an-
gle DBE to the angle in the segment BCD.

From the point B draw (11. 1.) BA at right angles to EF, and take
any point C in the arch BD, and join AD, DC, CB ; and because the
straight line EF touches the circle ABCD in the point B, and BA is
drawn at right angles to the touching line, from the point of contact
B, the centre of the circle is (10. 3.)
in BA ; therefore the angle ADB in
a semicircle, is a right (31. 3.) angle,
and consequently the other two angles,
BAD, ABD, are equal (32. 1.) to a
right angle ; but ABF is likewise a
right angle ; therefore the angle ABF
is equal to the angles BAD, ABD :
take from these equals the common
angle ABD, and there will remain
the angle DBF equal to the angle
BAD, which is in the alternate s^-
ment of the circle. And because ABCD is a quadrilateral figure in
a circle, the opposite angles BAD, BCD are equal (22. 3) to two right
angles ; therefore the angles DBF, DBE, being likewise equal (13.
1 .) to two right angles, are equal to the angles BAD, BCD ; and DBF

ELEMENTS

haa been proved equal to BAD : therefore the remaining angle DBE
is equal to the angle BCD in the alternate segment of the circle.
Wherefore, if a straight line, dsc. Q. E. D.

PROP. XXXIII. PROB.

Upon a given tiraight line to describe a segment of a eireUf coniaimng
an angle equal taa given rectilineal angle.

Let A.B be the given straight line, and the angle at C the given rec-
tilineal angle ; it is required to describe upon the given straight line
AB a segment of a circle, containing an angle equal to the angle C.

First, let the angle at C be a right angle ; bisect (10. 1.) AB in F,
and from the centre F, at the dis-
tance FB, describe the semicircle
AHB ; the hngle AHB being in a
semicircle is (31. 3.) equal to the
right angle at C.

But if the angle C be not a right
angle at the point A, in the straight
line AB, make (23. 1.) the angle BAD equal to the angle C, and from
thoypoint Adraw(ll. 1.) AE -rr *

at right angles to AD ; hi- .
sect (lO; 1.) AB in F, and

from F draw (11. 1.) FG at / >^JJ

right angles to AB, and join
GB : Then because AF is
equal to FB, and FG com-
mon to the triangles AFG, O i AV
BFG, the two sides AF, FG
are equal to the two BF,
FG ; but the ^ngle AFG is
also equal to the angle BFG ; -r\>
therefore the base AG is ,
equal (4. 1.) to the base GB f and the circle described from the cen*
tre G, at the distance GA, shl^l pass through the point B ; let this be
the circle AHB :^ And because' from the point A the extremity of the
diameter A£, AD is drawn at^ right
angles to AE, therefore AD (Cor.
16r 3.) touches the circle; and
because AB, drawn from the point
of contact A, cuts the circle, the
angle DAB is equal to the angle
in the alternate segment AHB
(32. 3.) ; but the angle DAB is
equal to the angle C, therefore
also the angle C is equal to the
angle in the segment AHB :

iyP GEOMETRY. BOOk lit.

\yherefbi'e, upon Ihe given straight line AB the seghient AHB of li
Circle is described which contains an angl6 equal to the giv^n angl^
kt C. Which Was to be done; »-

PROP. XJtXiV. PROfl. ^

'to cutoff a segment from a, given circle which shall contain mn dngU

equal to a given rectilineal angle*

Let ABC be th^ given circle, and D the given rectilineal angle ; it
is required to cut off a segment from the circle ABC that shaU con^
tain ati angle equal to the angle D.

Draw (17. 3.) the straight line £F touching the drcte ABC) in tiril
)>oint B, and at the point B, in the
{Straight line BF make (23. 1.)
the angle FBC equal to the angle
D ; therefore, because the straight
line £F touches the circle ABC,
dnd BC is drawn from the point
of contact B. the angle FBC is
equal (32. 3.) to the angle in the
isiUernate segment BAG ; but the
angle FBC is equal to the angle
D : therefore the angle in the
^segment 6AC is equal to the angle D : wherefore th^ i^^gmeht BAd
is cut off from the given circle ABC containing an angle equal to ihU
given angle D. Which was to be done;

PROP. XXXV. THEOk^

J(f itdo straight lines within a circle cut one another, the rectangle cofi:
tained by the segments of one of them is equal to the rettangle com
iained by the segments of the other. -

Let the two straight lines AC,BD,\vithin the circle ABCD, tiii tiM
hhother in the point E : the rectangle contdintd by AE, EC is equal
io the rectangle contained by BE, ED.

If AC, BD pass each of them through the
centre, so that E is the x^entre, it is evident
That AE, EC, BE, ED, being all equal, the
rectangle AE.EC is likewise equal to the
rectangle BE/ED.

- But let one of them BD pass through the
centre, and ciit the other AC, which dfoes nd(
pass through the centre, at right angles in the
point E ; then, if BD be bisected in F, F is (he centref of th6 ^mi^

ELEMENTS

ABCD ; join AF : and because BD, which passes through the centre^

cuts the straight line AC, which does not

pass through the centre at right angles, in

E, AE, EC are equal (3. 3.) to one another ;

and because the straight line BD is cut into

two equal parts in the point F, and into two

unequal in the point £, BE.ED (5. 2.) +

EP = FB" = AF'. But AF' = AE' +

(47. 1.) EP, therefore BE.ED + EF'' =

AE'+EP, and taking EF' from each, BE.

ED=AE«=sAE.EC.

Next, Let BD, which passes through the centre, cut the other AC»

which does not pass through the centre,
in E, but not at right angles; then, as be*
fore, if BD be bisected in F^ F is the centre
of the circle. Join AF, and from F draw
(12. 1.) FG perpendicular to AC ; there-
fore AG is equal (3. 3. ) to GC ; wherefore
AE.EC+(5. 2.) EG'= AG», and adding
GF'to both, AE.EC+EG'+GF*=AG«
+ GP. Now EG'+ GF' = EFS and
AG'+GF'= AF' ; therefore AE.EC+EF'= AP= FB». But FB?
=BE.ED+(5. 2.) EFS therefore AE.EC+EF'==BE.ED+EF',aBd
taking EF' from both, AE. EC = BE.ED.

Lastly, Let neither of the straight lines
AC« BD pass through the centre : take the
centre F, and through E, the intersection
of the straight lines AC, DB, draw the di-
ameter GEFH : and because, as has been
shown, AE.EC= GE.EH, and BE.ED =
GE.EH ; therefore AE.EC = BE.ED.
Wherefore, if two straight lines, &c.
Q. £. D.

PROP. XXXVL THEOR.

If from any point withovt a circle Itco straight lines he drawn, one of
which ciUs the circle^ and the other tottches it ^ the rectangle contain'- ^
■ ed by the wJiole line which cuts the circle, and the part of it without
the circle, is equal to the square of the line which Umches it.

Let D be any point without the circle ABC, and DC A, DB two
straight lines drawn from it, of which DCA cuts the circle, and DB
touches it: the rectangle AD.DC is equal to the square of DB.

OP GEOMETRY. BOOK III.

Either- DCA passes through the centre, or
it do6s not ; first, Let it pass through the cen-
tre E, and join EB ; therefore the angle
EBD is a right (16. 3.) angle : and because
the straight line AC is bisected in E, and pro-
duced to the point D, AD. DC + EC^ =
ED^ (6. 2.). But EC=EB, therefore AD. B
DC + EB^ = ED^ Now ED* = (47. 1.)
EB^ + BD^ because EBD is a right angle ;
therefore AD.DC+EB3~EB''4.BD^ and
taking EB^ from each, AD.DC=BD'.

But, if DCA does not pass through the
centre of the circle ABC, take (1. 3.) the
centre E, and draw EF perpendicular (12.
1.) to AC, and join EB, EC, ED: and be-
cause the straight li^ne EF, which passes
through the centre, cuts the straight line
AC, which does not pass through the cen-
tre, at right angles, it likewise bisects (3. 3.)
it ; therefore AF is equal to FC ; and be-
cause the straight line AC is bisected in F,
and produced to D (6. 2.), AD.DC+FC^= o
FD^ ; add jfflE^ to both, then AD.DC+FC^
4-FE'=FD^+FE^ But (47. 1.) EC2=FC*
+ FE^ and ED^ = FD^ + FE% because
DFE is a right angle ; therefore AD.DC+
EC'*=EI>*. Now, because EBD is a right
angle, ED»=EB^+BD2=EC2+BD^ and
therefore, AD.DC+EC^—EC^+BD", and
AD.DC=BD^

Wherefore, if from any point, &c. Q. E. D.

Cor. If from any point without a circle,
there be drawn two straight lines cutting it,
as AB, AC, jthe rectangles contained by the
whole lines and the parts of them without
the .circle, are equal to one another, viz. ,
BA.AE=CA.AF ; for each of these rect-
angles is equal to the aquare of the straight D
line AD, which touches the circle.

Wft

EI4EM6NTS, &C,

PROP. XXXVII. THEOR.

fffl^cm a p^int pithout a circle there he drawn tvx> straight lines, one
of which cuts the circle^ and4he other meets it ; if the rectangle con-*
tained hy the whole line^ which cuts the circle^ and the part of it wAh^
out the circle, be -equal to the sqUare of the line which meets it, the
line which meela shall touch the circle*

Let any point D be taken without the circle ABC, -and from it let
two straight lines DCA and DB be drawn, of which DCA cuts the
circle, and PB meets it \ if the rectangle AD.DC, bo equal to the
square of DB, DB touches the circle.

Draw (17. 3.) the straight line DE touching the circle ABC ; find
the centre F, and join FE, FB, FD; then FED is aright (18. 3.)
imgle : and because DE touches the circle ABC, and DCA cuts it^
|he rectangle AD.DC is equal (36. 3.) to the square of DE ; but the
fectangle AD.pC is, by hypothesis, equal to the square of DB 9
therefore the square of DE is equal to the square of DB ; and th»
straight line DE equal to the straight line DB : t|

l)ut FE is equal to FB, wherefore DE, EF are ^

^qual to DB, BF ; and the base FD is com-,
moijk to the two triangles DEF, DBF ; there^
fore the angle DEF is equal (8. I.) to the an->
^le DBF ; ^x^^ D£\F is a right angle, therefore
filse DBF is a right angle ; but F^ if pro- S^
^uced, is a diameter, and the straight line
which is drawn at right angles to a diameter,
from the extremity of it, touches (16. 3.) the
^ircle : therefore DB touches the circle ABC.
ilVherefoi^e^ if frgm a poiat, dfc. Q. E. D.

ELEMENTS

GEOMETRY.

* t

BOOK IV-

DEFINITIONS-

A RECTILINEAL figure is said to be inscribed ia another rectilineal
figure, when all the angles of the inscribed
figure are upon the sides of the figure in which
it is inscribed, each upon each*

II.

In like manner, a figure is said to be described
about another figure, when all the sides of the
circumscribed figure pass through the angular

points of the figure about which it is described, each through eachi

IIL

A rectilineal figure is said to be inscribed in
a circle, when all the angles of the inscrib-
ed figure are upon the circumference of
the circle.

IV.

A rectilineal figure is said to be described
about a circle, when each side o^the cir-
cumscribed figure touches the circumfer-
ence of the circle-

In like manner, a circle is said to be inscrib*
ed in a rectilineal figure, when the circum-
ference of the circle touch^ti each si<|e of
Ibe figuito,

94 ELEMENTS

VI.

A circle is said to be described about a recti-
lineal figure, when the circumference of
the circle passes through all the angular
points of the figure about which it is de^
scribed.

, VII.

A straight line is said to be placed in a circle, when the extremities
of it are in the circumference of the circle.

PROP. I. PROB.-

In a given circle to place a straight line equal to a given etraight line,
not greater tJum the diameter of the circle.

Let ABC be the given circle/ and D the given straight line, nol
greater than, the diameter of the circle.

Draw BC the diameter of the cir-
cle ABC ; then, if BC is equal to
D, the thing required is done ; for
in the circle ABC a straight line BC
is placed equal to D : But, if it is
not, BC is greater than D ; make
CE equal (3. 1.) to D, and from
the centre C, at the distance CE,
describe the circle AEF, and join

CA ; Therefore, because C is the tj

centre of the circle AEF, CA is

equal to CF ; but . D is equal to CE ; therefore D is equal to C A :
Wherefore,, in the circle ABC, a straight line is placed, equal to the
given straight line D, which is not greater than the diameter of the
circle. Which was to be done.

PROP. II. PROB.

Jn a given circle to describe a triangle equiangular to a given triangle.

Let ABC be the given circle, and DEF the given triangle ; it is
required to inscribe in the circle ABC a triangle equiangular to the
triangle DEF.

Draw (17. 3.) thB straight line GAH touchiogthe circle in the point
A, and at the point A, in the straight line AH, make (23. 1.) the angle
HAC equal to the an^le DEF ; and at the point A, in the straight lin&

OF GEOMETRY. BOOK IV.

AG^ make tlie angle GAB
equal to the angle DFE, and
join BC. Therefore, because
HA G touches the circle ABC,
and AC is drawn from the
point of contact, the angle
HAC is equal (32. 3.) to the
angle ABC in the alternate
segment of the circle : But
HAC is equal' to the angle ^
DEF ; therefore also the an-
gle ABC is equal to DEF ; '

for the same reason, the angle ACB^is equal to the angle DFE ; there-
fore the. remaining angle BAC is equal (32. 1.) to the remaining angle
EDF : Wherefore the triangle ABC is equiangular to the triangle
DEF, and it is inscribed in the circle ABC. Which was to be done.

PROP. m. PROB.

About a given circle to describe a triangle equiangular to a given

triangle.

Let ABC be the given circle, and DEF the given triangle ; it is re^
quired to describe a triangle about the circle ABC equiangular to the
triangle DEF.

Produce EF both ways to the points G, H, and find the centre
K of the circle ABC, and from itijraw any straight line KB ; at the
point K in the straight line KB, make (23. 1.) the angle BKA equal
to the angle DEG, and the angle BKC equal to the angle DFH ; and
through the points A, B, C, draw the straight lines LAM, MBN, NCL
touching (17. 3.) the circle ABC : Therefore, because LM, MN, NL
touch the circle ABC in the points A, B, C, to which from the centre
are drawn KA, KB, KC, the angles at the points A, B, C, are right
(18. 3.) angles. And because the four angles of the quadrilateral
figure AMBK are equal to four right angles, for it can be divided int^
two triangles; and because two of them,KAM, KBM^are right angles.

a E » H

to Elements

the other two A KB, AMB are equal to two right aDgles : Bat the M^

i^les DEG, DEF are likewise equal (13. 1.) to two right angles; there^
ore the aagles A KB, AMB are equal to the angles DEG, DEF, of
llrhich AKB is equal to DEG ; wherefore the remaining angle AMB id
6qual to the remaining angle DEF. In like manner, the angle LMN
may be demonstrated to he equal to DFE ; and therefore the remainr.
ing angle ML^ is equid (32. 1.) to the remaining angle EOF: Wherei
fore the triangle LMN is equiangular to the triangle DEF : And it iff
described about the circle ABC. Whieh was to be done^

PROP; IV. PROB.

To inscribe a circle in a given triangle.

. /Jjot the given triangle be ABC ; it is required to inscribe a circle'
in ABC.

Bisect (9. 1.) the angles ABC, BCA by the straight lines BD, €0
ineeting one another in the point D, from which draw (12. 1.) DE, DF/
i>G perpendiculars to AB, BC, CA.
Then because the angle EBD is equal
to the angle FBD, the angle ABC be^
ing bisected by BD ; and because the
right angle BED, is equal to the right
angle BFD, the two triangles EBD,
FBD have two angles of the one equal
to two angles of the other ; and the side
BD, which is opposite to one of the
equal angles in each, is common to
both ; therefore their other sides are
equal (26. 1.) ; wherefore DE is.equal
to DF. From the same reason,DG is equal to DF ; therefore the threes
straight lines DE, DF, DG, are equal to one another, and the circler
described from the centre D, at the distance of any of tbemj will
pass through the extremities of the other two, cmd will touch the
straight lines'AB, BC, CA, because the angles at the points E, F, G,
are right abgles, and the straight line which is drawn from the extre-
t&ity of a diameter at right angles to it, touches (Cor. 16. 3.) the cir^
cle. Therefore the straight lines AB, BC, CA, do each of then>
touch the circle, and the circle EFG is inscribed in the triangle ABC*
Which was to be done.

PROP. V. PROB^.

To describe a circle about a given triangle.

Let the given triangle be ABC ; it is required to describe a circle
about ABC.

Bisect (10. 1.) AB| AC in the points D, E, and from these pointsr*
*aw DF, EF nt right angles (.IL 1.) to AB, AC ; DF, EF produccii?

OF GEOMETRY. BOOK IV.

will meet one sanother ; for, if they do not meet, they are parallel^
wherefore, AB, AG, which are at right angles to them, are parallel,
which is absurd ; Let them meet in F, and join FA ; also, if the point
"F be not in BC, join BF, OF : then, because AD is equal to BD, and
*DF oemmon, andat right angles to AB, the base AF is equal (4. 1;)
to the base FB« In like manner, it may be shewn that OF is equal to .
^A ; and therefore BF is equal to F€ ; and FA, FB, FC are equal to
••Be another ; wherefore the circle described from the centre F, at the
distance of one of them, will pass through the extremities of the other
two, and be described about the triangle ABC, which was to be done.
Cob* When the centre of the circle falls within the triangle, each
of its angles is less than a right angle, each of them being in a seg-
ment greater than a~ semicircle ; but when the centre is in one of
dbe sides of the triangle, the angle opposite to this side, being in a se-
micircle, is a right angle : and if the centre falls without the triangle,
the. angle opposite to the side beyond which it is, being in a segment
less than a semicircle, is greater than a right angle. Wherefore, if the
given triangle be acute angled, the centre of the circle falls within
it : if it be a right Imgled triangle, the centre is in the side opposite
to the right angle ; and if it be an obtuse angled triangle, the centre
falls mthotttthe triangle, beyond the side opposite to the obtuse angle.

PROP. VI. PROB, ^
To inscribe a square in a given circle.

Let ABCD be the given circle ; it is required to inscribe asquai^e
inABCD.

Draw the diameters AC,BD at right angles to one another, and join
AB, BC, CD, DA ; because B£js equal to ED, E being ihe centre,
and because E A is at right angles to B0, a

and common to the triangles ABE, ADE ;
the base BAis equal (4. 1.) to the base
AD ; and, for the sam^ reason, BC, CD
are each of them equal to BA or AD ;
' therefore, the quadrilateral f^ure ABCD |B
• is equilateral. It is also rectangular;
{sr the straight line BD heing a diameter
.of the circle ABCD, BAD is.as^m^cir*
ele ; wherefojre^the angle BAPis & ;^ight
(31. 3.) angle ; for the same reason each

ELEMENTS

of the angles ABC, BCD, CDA is a right angle ; therefore the qnad*
rilateral figure ABCD is rectangular, and it has heen shewn to be
equilateral ; therefore it is a square ; and it is inscribed in the circle
ABCD. Which was to be done.

PROP. VII. PROB.

To describe a square ahmU a given circle.

Let ABCD be the given circle ; it is required to describe a square
about it.

Draw two diameters AC, BD of the circle ABCD, at right angles
to one another, and through the points A, B, C, D draw (17. 3.) FG,
GH, HK, KF touching the circle ; and because FG touches the cir-
cle ABCD, and £A is drawn from the centre E to the point of con-
tact A, the angles at A are right (16. 3.) angles ; for the same rea-
son, the angles at the points B, C, D are right angles ; and because
the angle AEB is a right angle, as likewise is EBG, GH is parallel
(26. 1.) to AC ; for the same reason,
AC is parallel to FK, and in like man-
ner, GF, HK may each of them b^ de-
monstrated to be parallel to BED ;
therefore the figures GK, GC, AK, FB,
BK are parallelograms ; and GF is
therefore equal (34. 1.) to HK, and
GH to FK ; and because AC is equal to
BD, and also to each of the two GH,
FK ; and BD to each of the two GF,
HK : GH, FK are each of them equal
to GF or HK ; therefore the quadrilateral figure FGHK is equilate-
ral. It is also rectangular ; for GBEA being a parallelogram, and
AEB a right angle, AGB (34. 1.) is likewise a right angle: In the
same manner, it may be shewn that the angles at H, K, F are right
angles ; therefore the quadrilateral figure FGHK is rectangular ; and
it was demonstrated to be equilateral ; therefore it is a square ; and
it is described about the cifcle ABCD. Which was to be done.

PROP. VIII. ^ PROB.

To inscribe a circle in a given square. ^

Let ABCD be the friven square ; it is required to inscribe a circle
in ABCD.

Bisect (10. 1.) each of the sides AB, AD, in the points F, E, and
through £ draw (31. 1.) EH parallel to AB or DC, and through F
draw FK parallel to AD or BC ; therefore each of the figures, AK,
KB, AH, HD, AG, GC/BG, GDis a parallelogram, and their oppo-
site sides are equal (34. 1.) ; and because that AD is c^ual to AB,-

OP GEOMETRY. BOOK IV-

and that AE is the half of AD, and AF the half of AB, AE is equal to
AF ; wherefore the sides opposite to these are equal, viz. F6 to
GE ; ia the same manaer it may be demonstratedi that GH, GE, are
ei^ch of them equal^to FG or GE ; there- a JJ jy

fore the four straight lines, GE, GF, GH, '~
GK, are equal to one another; and the
circle described from the centre G, at the
distance of one of then^ will pass through
the extremities of the other three ; and
will also touch the straight lines AB, BQ
CD, DA, because the angles at the points
E, F, H, K, are right (29. 1.) angles, and
because the straight line which is 'drawn
from the extremity of a diameter at right
angles to it, touches the circle (l^:^3i) ; therefore each of the straight
lines AB,BC, CD, DA touched the circle, which is therefore inscrib-
ed in the squares ABCD. Which was to be done.

PROP. IX. PROB.

To describe a cirde about a given square*

Let ABCD^be the given square ; it is required to describe a circle
about it.

Join AC, BD, cutting one another in E ; and because DA Js equal
to AB, and AC common to the triangles DAC, BAC, the two sides
Pa, AC are equal to the two BA, AC, and the base DC is equal to
the base BC ; wherefore the angle DAC is
equal (8. 1.) to the angle BAC, and the an-
gle DAB is bisected by the straight line AC.
In the same manner it may be demonstrate
ed, that the angles ABC, BCD, CDA are
severally bisected by the straight lines BD,
AC ; therefore, because the angle DAB is
equal to the angle ABC, and the angle EAB
is the half of DAB, and EBA the half of
ABC ; the angle EAB is equal to the angl&
EBA : and the side E A (6. 1.) to the side EB. In the same manner,
it may be demonstrated, that the straight lines EC, ED are each 4>f
them equal to EA, or EB ; therefore the four straight lines EA, EB,
EC, ED, are equal to one another ; and the circle described from the
centre E, at the distance of one of them, must pass through the extre-
mities of the other three, and be described about the square ABCD.
Which was to be done. .

■J , .

109 ELEMENTS

PROP; X, PRQB.

To describe an isosceles triangle^ hating 4ach of the angles ai the ham

double of the third angle.

Take any straight line AB/and divide (II4 2.) it ih the point C, so
that the rectangle AB.BC may be equal to tjbe square of AG ; and
from the centre A, at the distance AB, describe the circle BDE, ia
which place (1. 4.) the straight line BD equal to AC, which is tet
greater than the diameter of the circle BDE ; j<»n DA> DC, and
about the triangle ADC describe (5. 4.) the circle ACD ; thfe trian*
gle ABD is such as is required, that is, each of the angles ABD, ADB
is double of the angle BAD.

Because the rectangle AB.BC is equal to the square of AG, and
AC equal to BD, the rectangle
AB.BC is equal to the square of
BD ; and because from the point
B without the circle ACD IWo
straight lines BCA, BD are drawn
to the circumference, one of
which cuts, and the other meets
the circle, and the rectangle AB. '
BC contained by the whole of tho
cutting line, and the part of it
without the circle, is equal to the
square of BD, which meets it ;
the straight line BD touches (37. •
3. ) the circle ACD. And because
BD touches the circle, and DC

is drawn from the point of contact D, the angle BDC is equal
(32. SA to* the angle DAC in the alternate segment of the circle, to
each or these add the angle CD A ; therefore the whole angle BDA
is equal to the two angles CDA, DAC ; but the exterior angle BCD j

is equal (32« 1.) to the angles CDA, DAC ; therefore also Bt)A is I

equal to BCD ; but BDA is equal (5. 1.) to CDB, because the side
AD is equal to the side AB ; therefore CBD, or -DBA is equal to
BCD ; and consequently the three angles BDA, DBA, BCD, are
equal to one another. And because the angle DBC is equal to the an-
gle BCD, the side BD is equal (6. 1.) to the side DC; but BD was made
equal to CA ; therefore also CA is equal to CD, and the angle CDA
equal (5. 1.) to the angle DAC ; therefore the angles CDA, DAC to-
gether, are double of the angle DAC ; but BCD is equal to the an*
gles CDA, DAC (32. 1.) ; therefore also BCD is double of DAC,
But BCD is equal to each of the angles BDA, DBA, and therefore
each of the angles BDA, DBA, is double of the angle DAB ; where-'
fore an isosceles triangle ABD is described, having each of the an-
gles at the base double of the third angle. Which was to be done*

B^ D

. OF GEOMETRY. BOOK IV. W,

' ^* €oft. I* The angle BAD is the £fth paH of two rigM angleB.
** For since each of the angles ABD and ADB is equal to twice the
^ angle BAD^ they are together equal to four times BAD» and there-
** fore all the three angles ABD, A£)B, BAD, taken together, are
** equal to five times the angle -BAD. But the three angles ABD,
*^ ADB, BAD are equal to two right angles, therefore five times the
** angle BAD is equal to two right angles ; or BAD is the fifUi part
of two right angles."

Cor. 2« Because BAD is the fifth part 6f two, or the tenth part
of four right angles, all the angles about, the centre A are together
** eqjual to ten times the angle BAD, and may therefore be divided*
*^inte ten parts each equal to BAD* And as these ten equal angles
" at the centre, must stand on ten equal arches, therefore the arch
*^ BD is one-tenth of the circumference f and the straight line BDy
<^that is, AC, is therefore equal to the side of an equilateral decagon
*' inscribed in the circle BDE,"

PROP. XI. PROB.

To inscribe an equUaterdl and equiangular pentagon in a given circle*

Let ABODE! be the given circle, it is required to inscribe aneqiu*^
literal and equiangular pentagon in the circle ABODE.

Describe (10. 4.) an isosceles triangle FGH, having each of the
angles at G, H, double of the angle at F ; and in the circle ABODE
inscribe (2. 4.) the triangle AOD eq^iiangular to the triangle FGH,
so that the angle OAD be equal to the angle at F, and each of the an*
^ gles AOD, ODA equal to the
angle at G or H ; wherefore
each of'the-angles AOD,CDA
is double of the angle OAD.
Bisect (9. 1.) the angles AOD,
CD A by the straight lines OE,
TDB ; and join AB, BO, DE,
£:A. abode is the penta-
gon required.

Because the angles AOD, gT
CD A are each of them double
of OAD, and are bisected by the straight lines OE, Dfi, the five angles
DAO, AOE, EOD, ODB, BDA are equal to one another ; but equal
angles stand upon equal (26. 3.) arches; therefore ^the five arches
AB, BO, OD, DE, EA are equal to one another ; and equal arches
are subtended by equal (29. Ss ) straight lines ; therefore the five
straight lines AB, BO, OD, DE, EA are equal to one another.
Wherefore the pentagon ABODE is equilateral. It is also equiangu-
^ lar ; because the arch AB is equal to the arch DE ; if to each be add-
' ed BOD, the whole ABOD is equal to the whole EDOB ; and the &ngle
AJSD stands on the arch ABOD, and the angle BA£ on the arch

l(tt ELEMENTS

EDCB : therefore the angle BAE is equal (27. 3.) to the angle.
AED : for the same reason, each of the angles ABC, BCD, CDE iti
equal to the angle BAE or AED : therefore the pentagon ABCDE
is equiangular : and it has been shewn that it is equilateral. Where-
fore, in tbs given circle, an equilateral and equiangular pentagon has
been inscribed. Which was to be done.

Otherwise.

*' Divide the radius of the given circle, so that the rectangle con*
** tained by the whole and one of the parts may be equal to the square
" of the other (11. 2.). Appiy in the circle, on each side of a given
" point, a line equal to the greater of these parts ; then (2. Cor. 10,
*'4.), each of the arches cut off will be one-tenth of the circumfer-
*' ence, and therefore the arch made up of both will be one*fifth of
** the circumference ; and if the straight line subtending this arcli be
** drawn, it will be the side of an equilateral pentagon inscribed in
" the circle."

PROP. XII. PROB.

To describe an equilateral and equiangular pentagon about a given drde.

Let ABCDE be the given circle, it is required to describe an 'equi-
lateral and equiangular pentagon about the circle ABCDE.

Let the angles of a pentagon, inscribed in the circle, by the last pro*
posidon, be in the points A, B, C, D, E, so that the arches AB, BC,
CD,^DE, £A are equal (11.4.); and through the points A, B, C, D/
E, draw GH, HK, KL, LM, MG, touching (17. 3.) the circle; take
the centre F, and join YE^ FK, FC, FL, FD. And because the straight
line KL touches the circle ABCDE in the point C, to which FC is
drawn from the centre F, FC is perpendicular (18. 3.) to KL ; there-
fore each of the angles at C is a right angle : for the same reason, the
angles at the points B, D are right angles; and because FCK is a right*
angle, the square of FK is equal (47. 1.) to the square of FC, CK.
For the same reason, the square of FK is equal to the squares of
FB, BK : therefore the squares of FC, CK are equal to the squares
of FB, BK, of which the square of FC is equal to the square of
FB ; the remaining square of CK is therefore equal to the remaining
square of BK, and the straight line CK equal to BK : and because
FB is equal to FC, and FK common to the triangles BFK, CFK, the
two BF, FK are equalto the two CF, FK ; and the base BK is equal
to the base KC ; therefore the angle BFK is equal (8. 1.) to the an-
gle KFC, and the angle BKF to PKC ; wherefore the angle BFC is
double of the angle KFC, and BKC double of FKC : for the same
reason, the angle CFD is double of the angle CFL, and CLD double
of CLF : and because the arch BC is equal to the arch CD, the an-
gle BFC is equal (27. 3.) to the angle CFD ; and BFC is double of
the angle KFC, and CFD double of CFL ; therefore the angle KFC is

OF GEOMETRY. BOOK !¥•

108

•qual to the angle CFL; now the
right angle FCK is equal to the right
angle FCL ; and therefore, in the
two triangles FKC, FLG, there are
two angles of one equal to two an-TTY
gles of the other, each to each,^ and-^^
the side FG, which is adjacent to
the equal angles in each, is common
to both ; therefore the other sides
are equal (26. 1.) to the other sides,
and the third angle to the third an-
gle ; therefore the straight line KG is

equal to CL, and the angle FKC to the angle FLO : and because EC
18 equal to CL, KL is double of KG ; in th^ same manner, it may be
shewn that HK is double of BK ; and because BK is equal to KG, as
was demonstrated, and KL is double of KC, and HK double of BK,
HK is equal to KL ; in like manner, it may be shewn that GH, GM,
ML are each of them equal to HK or KL : therefore the pentagon
GHKLM is equilateral. It is also equiangular ; for, since the angle
FKC is equal to the angle FLG, and the angle HKL double of the
angle FKC, and KLM double of FLG, as was before demonstrated,
the angle HKL is equal to KLM ; and in like manner it maybe shewn,
that each of the angles KHG, HGM, GML is equal to the angle
HKL or KLM ; therefore the five angles GHK, HKL, KLM, LMG,
MGH being equal to one another, the pentagon GHKLM is equian-
gular ; and it Is equilateral as was demonstrated ;. and it is described
about the circle ABGDE. Which was to be donQ.

PROP. Xllh PROB.

To inscribe a circle in a given equUtxterdl and equiangtdar pentagon^

Let ABCDE be the given equilateral and equiangular pentagon ;
it is required to inscribe a circle in the pentagon ABCDE.

Bisect (9. 1.) the angles BCD, GDE by the straight lines CF, DP,
and from the point F,in wliich they meet, draw the straight lines FB,
FA, F£ ; therefore, since BG is equal to CD, and GF common to the
triangles BGF, DGF, the two sides BG, GF are equal to the two DC,
GF; and the angle BGF is equal to the angle DCF : therefore the base
BF is equal (4. 1.) to the base FD, and the other angles to the other
angles, to which the equal sides are opposite ; therefore the angle CBF
is equal to the angle GDF : and because the angle CDE is double of
CDF, and GDE equal to GBA, and GDF to GBF ; GBAis also double

104

ELEMENTS

of the angle CBF ; therefore the
angle ABF is equal to the angle
CBF ; wherefore the angle ABC is
bisected by the straight line BF : In
the same manner, it may be demon. _
strated that the angles BAE, AED, -B'
are bisected by the straight lines
AF, EF : from the point F draw
(12, 1.) FG, FH, PK, FL, FM per-
pendiculars to the straight lines ABy
BC, CD, DE, EA : and because the
,angle HCF is equal to KCF,and the
Tight angle FHC equal to the right angle FKC ; in the triangles FHC,
FKC there are two angles of one equal to two angles of the other, and
the side FC, which is opposite to one of the equal angles in each, is
common to both ; therefore, the other sides shall be equal (26. 1.),
'each to each ; wherefore the perpendicular FH is equal to the perpen-
dicular FK: in the same manner it may be demonstrated, that FL, FM,
FG are each of them equal to FH, or FK ; therefore the five straight
lines FG,.FH, FK, FL, FM are equal to one another ; wherefore the
'circle described from the centre F, at the distance of one of these five,
will pass through the extremities of the other four, and touch the
straight lines AB, EC, CD, DE, EA, because that the angles at the
^points G, H, K, L, M are right angles, and that a straight line drawn
'from the extremity of the diameter of a circle at right angles to it,
touches (Cor. 16. 8.) the circle ; therefore each of the straight lines
AB, BC, CD, DE, E A touches the circle ; wherefore the circle Is
inscribed in the pentagon ABCDE. Which was to be done*

PROP. XIV- PROB.

Tode9eribe actrde about a given equUaterdtandrequiangtdar

pentagon,

Let ABCDE be the given equilateral and equiangular peniagoa;
it is required to describe a circle about it«

Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, FD,
and from the point F, in which they a

meet, draw the straight lines FB, FA,
F£ to the points B, A, E. It may he
demonstrated, in the same manner as in
'4he preceding proposition, that the an- 3i
gles CBA, BAE, AED are bisected by
the straight lines FB,FA, FE : and be.
cause that the angle BCD is equal to
the angle CDE, and that FCD is the
half of the angle BCD, and CDF the _.

half of CDE ; the angle FCD is equal C

OP GfiOMfeTRt. BOOK IV.

10&

•

lo PDC ; wherefore the side CF is eqaal (6. 1.) to the side FD : In
like manner it may be demonstrated, that FB, FA, FE are each of
ihem equal to FC, or FD : therefore the five straight lines FA, FB,
FC, FD, FE are equal to oile another ; and the circle described from
the centre F, at the distance of one of them, will pass through the
extremities of the other four, and foe described about the equilateral
and equiangular pentagon ABODE. Which was to be done.

PROP. XV^ PROB.

. To inscribe an equilateral and equiangular hexagon in a green circle*

Let ABCDEF be the given circle ; it is required to inscribe an
equilateraland equiangular hexagon in it.

Find the centre G of the eircle ABCDEF, and draw the diameter
AGD : and from D, as a centre, at the distance DGy describe the
circle EGCH, join EG, CG, and produce them to the points B, F ;
end join AB, BC, CD, DE, EF, FA: the hexagon ABCDEF is
equilateral and equiangular.

Because G is the centre of the circle ABCDEF, GE is equal to
GD : and because D is the centre of the circle EGCH, DE is equal
to DG ; wherefcHre GE is equal to ED^ and the triangle EGD is equi-
lateral ; and therefore its three angles EGD, GpjE> DJSG are equal
to one another (Cor. 5. 1.) ; and the three angles of a triangle arQ
equal (32> 1.) to two right angles ; therefore the angle EGD is the
third part of two right angles : In the a

same manner it may be demonstrated that
\he angle DGC is also the third part of
two right angles : and because the straight £^
line GC diakes. with EB the adjacent an-
gles EGC, CGB equal (13. l.jf to two
right angles : the remaining angle CGB is
the third part of two right angles : there-
fore the angles EGD, DGC, CGB, are
equal to one another ; and also the angles
vertical to them, BGA, AGF, FGE (i5.
1.) ; therefore the six angles EGD, DGC,
CGB, BGA, AGF, FGE are equal to one
another. But equal angles at the centre
«tand upon equal (26. 3.) arches : there*
fore the six arches AB, BC, CD, DE, EF,
FA are equal to one another : and equal arches are subtended by
equal (29. 3.) straight lines; therefore the six straight lines are
equal to one another, and the hexagon ABCDjBF is equilateral. It
is also equiangular; for, since the arch AF is equal to ED, to each of
these add the arch ABCD ; therefore the whole arch FABCD shall
be equal to the whole EDCBA : and the angle FED stands upon the
arch FABCD, and the angle AFE upon EDCBA; therefore the angle

106 ELEMENTS, <kc.

AFE is equal to FED : io the same manner it may b^ demonfltrated^
that the other angles of the hexagon ABCDEF are each of them
equal to the angle AFE or FED ; therefore the hexagon is equian.
gular ; it is also equilateral, as was shown ; and it is inscribed in ^
the given circle ABCDEF. Which was to be done.

Cor. From this it is manifest, that the side of the hexagon is equal
to the straight line from the centre, that is, to the radius qf the circle.

And if through the points A, B, C, D, E, F, there be drawn straight
lines touching the circle, an equilateral and equiangular hexagon
shall be described about it, which may be demonstrated from what
has been said of the pentagon ; and likewise a circle may be inscrib.'
ed in a given equilateral and equiangular hexagon, and circumscrib-
ed about it, by a method like to that used ibr the pentagon.

PROP. XVI. PROB.

To inscribe an eqwlaieral and equiangtdar quindecagon in a given

circle.

Let ABCD be the given circle ; it is required to inscribe an equi-
lateral and equiangular quindecagon in the circle ABCD.

Let AC be the side of an equilateral triangle inscribed (2. 4.) in
the circle, and AB the side of an equi-
lateral and equiangular pentagon inscrib-
ed (11. 4.) in the same; therefore, of
such equal parts as the whole circum-
ference ABCDF contains fifteen, the S^
arch ABC, being the third part of the
whole, contains five ; and the arch AB, £|
which is the fiflh part of the whole,
contains three ; therefore BC their dif- <
ference contains two of the same parts :
bisect (30. 3.) BC in E ; therefore
BE, EC are, each of them, the fifteenth

part of the whole circumference ABCD : therefore, if the straight
lines BE, EC be drawn, and straight lines equal to them be placed
(1. 4.) around in the whole circle, an equilateral and equiangular
quindecagon will be inscribed in it. Which was to be done.

And in the same manner as was done in the pentagon, if through
the points of division made by inscribing the quindecagon, straight
lines be drawn touching the circle, an equilateral and equiangular
quindecafgon may be described about it : And likewise, as in the
pentagon, a circle may be inscribed in a given equilateral and
equiangular quindecagon, and circumscribed about it.

SlLEMENTS

GEOMETRY.

BOOK V.

In the demonstrations of this book there are ceTioiik*^ signs or
characters'*^ which it has been found convenient to employ.

«<

^' 1. The letters A, B, C, &c. are used to denote magnitudes of any
kind. The letters m^ n, p, 9, are used to denote numbers only.

"2. The. sign + (plus), written between two letters, that denote
magnitudes or numbers, signifies the sum of those magnitudes or
'< numbers* Thus, A + B is the sum of the two magnitudes denoted
*' by the letters A and B ; m-\-n is the sum of the numbers denoted
" by m and n. .

" 3. The sign — (minus), written between two letters, signifies the
" excess of the magnitude denoted by the first of these letters, which
'^ is supposed the greatest, above that which is denoted by the other.
^'Thus, A— B signifies the excess of the magnitude A above the
"*' magnitude B.' ,

" 4. When a number, or a letter denoting a number, is written
*' close to another letter denoting a magnitude of any kind, it sig-
^< nifies that the magnitude is multiplied by the number. Thus,
<< 3A signifies three times A ; mB, m times B, or a multiple of B by
" m. When the number is intended to multiply two or more mag-
<* nitudes that follow, it is written thus, m(A+B), which signifies
^< the sum of A and B taken m times ; m(A-'B) is m times the ex-
** cess of A above B.

" Also, when two letters that denote numbers are written close to
** one another, they denote the product of those numbers, when mul-
" tiplied into one another. Thus, mn is the product of m into n ; and
^^ rank, is A multiplied by the product of m into ». ^

** 5, The sign = signifies the equality of the magnitudes denoted

ld8 ELEMENTS

*< by the letters that stand on the opposite sides of it ; A^^B signifies
" that A is equal to B ; A+B=C — D signifies that the sum of A and
^ B is equal to the excess of C above D.

" 6. The sign 7 is used to signify that the magnitudes between
** which it is placed are unequal, and that the magnitude to which the
*^ opening of the lines is turned is greater than the other. Thus A
^* 7 B signifies thiit A is greater than B ; and A Z B signifies thai
« A is less than B.**

DEFINITIONS.

A less magnitude is said to he a part of a greater magnitude, when
the less measures the greater, that is, when the less is contained
a certain number of times, exactly^ in the greater.

A greater magnitude rs said to be a raultipre of a less, when the
• greater is measured by the less, that is, when the greater contakv?
the less a certain number of times exactly.

III.

Satio is a mutual relation of two magnitudes, of the same kind, to one
another, in respect of quantity.

IV.

Magnitudes are saicT to be of the same kind, when the less can be mul-
tiplied so as to exceed the greater ; and it is only such magnitudes
that are said to hare a ratio to one another^

■ v..

If there be four magnitudes, and if any equimultiples whatsoever be
taken of the first and third, and any equimultiples whatsoever of
the second and fourth, and if, according as the multiple of the first
is greater than the multiple of the second, equal to it, orless,. the
multiple of -the third is also greater than the multiple of the fourth,
equal to it, or less ; then the first of the magnitudes is saidlo have
to the second the same ratio that the third has ta the fourth.

VI.

Magnilud-es are said to be proportionals^ when the first has the same-
ratio to the second that the thM has to the fourth ; and the third to*
the fourth the same ratio which the fiflh has to the sixth, and so ott
whatever be their number.

« When four magnitudes, A, B, €, D are proportionals, it is usual to^
« say that A is to B as C to D, and to write them thus„ A : B ::
•* C I D,,or thus„ A ^B=G I IX'*

OF GEOMETRY. BOOK.V. lOi

VIL ' .

When of the equimultiples of four magnitudes, taken as in the fifth
definition, the multiple of the first is greater than that of the se-
cond, but the multiple of the third is not greater than the multiple
of the fourth : then the first is said to have to the second a greater
ratio than the third magnitude has to the fourth : a^d, on the con-
trary, the third is said to have to the fourth a less ratio than the
first has to the second.

VIII.

When there is any number of magnitudes greater than two, of which
the first has to the second the same ratio that the second has to the
third, and the second to the third the same ratio which the third
has to the fourth, and so on, the magnitudes are said to be conti-
nual proportionals.

IX.

When three magnitudes are continual proportionals, the second is said
to be a mean proportional between the other two.

When there is any number of magnitudes of the same kind, the first
is said to have to the last the ratio compounded of the ratio which
the first has to the second, and of the ratio which the second has
to the third, and of the ratio which the third has to the fourth, and
so on unto the last magnitude.

For example, if A , B, C, D, be four magnitudes of the isame kind, the
first A is said to have to the last D, the ratio compounded of the
ratio of A to B, and of the ratio of B to C, and of the ratio of ta
D ; or, the ratio of A to D is said to be compounded of the ratios
of A to B, B to C, and C to D.

And if A : B : : E : F ; and B : C : : G : H, ^d C : D : : K : L,
then, since by this definition A has to D the ratio compounded
of the ratios of A to B, B to C, C to D; A may also be said to have
to D the ratio compounded of the ratios which are the same with
the ratios of E to F, G to H, and K to L.

In like maniier, the same things being «upposedy if M has to N the
same ratio which A has to D, then, for shortness' sake, M is said
to have to N a ratio compounded of the same ratios which com.
pound the ratio of A to D ; that is, a ratio compounded of the ratios
of E to F, G to H, and K to L.

XI.

If three magnitudes are continual proportionals, the ratio of the first
to the third is said to be duplicate of the ratio of the firstto these.
con4*

*' Thus, if A. be to B as B to C, the ratio of A to C is said to be dupli-
" cate of the ratio of A to B. Hence, since by the last definition^
*< the ratio of A to C is compounded of the ratios of A to B, and B

1 10 ELEMENTS

*^ to Cy a ratio, which is compounded of two equal ratios, is du-
" plicate of either of these ratios."

XII.

If four magnitudes are continual proportionals, the ratio of the first to
the fourth is said to be triplicate of the ratio of the first to the ise-
cond, or of the ratio of the second to the third, &c.
So also, if there are five continual proportionals ; the ratio of the
first to the fifth is called quadruplicate of the ratio of the first to
the second ; and so on, according to the number of ratios. Hence,
^ a ratio compounded of three equal ratios, is triplicate of any
" one of those ratios ; a ratio compounded of four equal ratios
" quadruplicate," &c.

XIII.

In proportionals, the antecedent terms are called homologous to one
another, as also the consequents to one another.

Geometers make use of the following technical words to signify cer-
tain ways of changing either the order or magnitude of propor-
tionals, so as that they continue still to be proportionals.

XIV.

Permutando, or alternando, by permutation, or alternately ; this word
is used when there are four proportion als, and it is inferred, that
the first has the same ratio to the third which the second has to
the fourth ; or that the first is to the third as the second to the
fourth : See Prop. 16. of this Book.

XV.

Invertendo, by inversion : When there are four proportionals, and it
is inferred, that the second is to the first, as the fourth to the third.
Prop. A. Book 6.

XVL

Componendo, by composition : When there are four proportionals,
and it is inferred, that the first, together with the second, is to the
second as the third, together with the fourth, is to the fourth. 18th
Prop. Book 5.

XVII.

Dividendo, by division : when there are four proportionals, and it is
inferred that the excess 'of the first above the second, is to the se-
cond, as the exeess of the third above the fourth, is to the fourth.
17ih Prop. Book 5.

XVIII.

Gonvertendo, by conversion ; when there are four proportionals, and
it is inferred, that the first is to its excess above the second, as the
third to its excess above the fourth. ' Prop. D. Book 5.

OP GEOMETRY. BOOK V. lU

XIX*

Ex sequali (sc. distantia), or ex s^quo, from equality of distance ; when
there is any number of magnitudes moife than two, and as many
others, so that they are proportionals when taken two and two of
each rank, and it is inferred, that the first is to the last of the first
rank of magnitudes, as the first is to the last of the others ; Of this
there are the two following kinds, which arise from the different
order in which the magnitudes are taken two and two.

XX.

^ £x sequali, from equality ; this term is used simply by itself, when the
first magnitude is to the second of the first rank, as the first to the
second of the other rank ; and as the second is to the third of the
first rank, so is ijie second to the third of the other ; and so on in
order, and the inference is as mentioned in the preceding defini*
tion ; whence this is called ordinate proportion.

It is demonstrated in the 22d Prop. Book 5.

XXI.

Ex aBqiiali, in proportione perturbata, sen inordinata : from eqnalityv
in perturb^te, or disorderly proportion ; this term is used when the
first magnitude is to the second of the first rank, as the last but one
is to the last of the second rank ; and as the second is to the third
of the first rank, so is the last but two to the last but one of the se-
cond rank ; and aa the third is to the fourth of the first rank, so iar
the third from the last, to the last but two, of the second rank ; and
so on in a cross, or irwerse^ order ; and the inference is as in the
19th definition. It is demonstrated in the 23d Prop, of Book 5«

AXIOMS.
I.

EaiTiMULTiFLES of the same, or of equal magnitudes, are equal to one
another.

II.

Those magnitudes of which the same, or equal magnitudes, are equi-
multiples, are equal to one another.

III.

A multiple of a greater magnitude is greater than the same multiple
of a less.

IV.

That magnitude of which a multiple is greater than the same multi-
pie of another, is greater than that other magnitude.

112 ELEMENtS

PROP. I. THEOR.

If any number of nuxgniludes he equi^tdtiples of as many other i^ each
6f each, vjJuft multiple soever any one of the first is of its part, the
same multiple is the sum qfaU the first of the sum of all the rest.

Let any number of magnitudes A, B, and C be equimultiples of as
many others, D, E, and F, each of each, A+B+C is the same mul-
tiple of D,-t-E+F, that A is of l3.

Let A contain D, B contain £, and C contain F, each the same

number of times, as, for instance, three times.

Then> because A contains D three times, A=5=D+E>4-t>«

For the same reason, B==?=E+E+E ;

And also, C = F+F+F.

Therefore, adding equals to equals (Ax. 2. 1.), A+B+C is equal
to D+E+F, taken three times. In the same manner, if A, B, and C
were each any other equimultiple of D, E, and F, it would be shown
that A+B+C was the same multiple of D+E+F. Therefore, &c.
Q. E. D. .

Cor. Hence, if m be an^y number, mD+mE+mF=ffi(D+E+
F). for mD, wiE, and «iF are multiples of D, E, and F by m, there-
fore their sum is also a multiple of D+E+F by m.

PROP. II. THEOR..

if to a multiple of a magnitude by any number, a multiple of the same
magnitude by any number be added, the sum unll be the same mulli*
pU of thai magnitude that the sum of the ttno numbers is ofunHy.

Let A=otC, and B=«C; A+B=:(m+n)C.

For, since A=mC, A=C+C+C+<&c. C being repeated m times.
For the same reason, B=C+C+ &c, C being repealed n times.
Therefore, adding equals to equals, A+B is equal to C taken m+n
times ; that is, A+B=(m + n)C. Therefore A+B contains C as oft
as the_re are units in m+n. Q. E. D.

Cor. 1. In the same. way, if there be any number of multiples
whatsoever, as A=mE, B=nE, C=pE, it is shown, that A+B+C=
{m+n+p)E.

Cor. 2. Hence also, since A+B+C=(m+n+|>)E, and since A
=»iE, B=^nE, and C^pE, mE+nE+jpE==(m+n+p)E.

PROP. ni. THEOR.

If the first of three magnitudes contain the second as oft as there are units
in a certain nttfkjber, and if the second contain the third also, as often
as there are units in a certain number, the first will contain the third
as of t as there are units in the product qftJiese two numbers.

Let A=mB, and B=nC ; then A=5»^nC.

OP GEOMETRY, BOOK V.

118

Since B^nC, mB=fiC+nC 4- &c. repeated m times. But nC+nC
&c« repeated m times is equal to C (2. Cor. 2. 5.), multiplied by n+n
+^G, ft beinff added to itself m times ; but n added to itself m times^
M ft multiplied by m, or imi. Therefore nC+nC+d^* repeated m
tiHiesssflinC ; wheoce also mB=iiiiiC, and by hypothesis AnmB,
Ihdrefore A^nmC. Thereibre, dsc. Q. £. D. \^

PROP. IV. THEOR. *

JjT the first of four magnitudes Juis the same ratio to the second which the

. third has to the fourth, and if any equimultiples whatever betaken of

ihefirsi and third, and any whaiever of the second and fourth ; the

muUiple of the first shall have the same ratio to the muUiple of the

second, that the muUiple of the third has to the multiple of the fourth.

♦

Let A : B : : G : Dy and let m and it be any two numbers ; mA :
nB : : mC : nD.

Take of mA and mC equimultiples by any number p, and of nB and
itD equimultiples by any number q. Then the equimultiples of mA,
and utO by p^ are equimultiples also of A and C, for they contain A
and C as oil as there are units in pm (3. 5.), and are equal to pmA and
|mrC. For the same reason the multiples of nB and nD by q, are ^B,
^D. Since, therefore, A : B : | G : D, and of A and G there are
taken any equimultiples, viz. jpmA and pmQ, and of B and D, any
equimultiples ^nB, ^D, if pmk be greater than ^nB, pmC must be
greater than qnD ^def. 5, 5.) ; if equal, equal ; and if less, less. But
|>mA, pmC are also equimultiples of mA and mC, and ^nB, ^nD are
equimultiples of nB and nD, therefore (def. 5. 5.}, mA : nB : : mC :
nD. Therefore, dec. Q. E. D.

Cor* In the same manner it may be demonstrated, that if A : B : :
C : D, and of A and C equimultiples be taken by any number m, viz*
mA and mC, mA : B : : mC : D. This may also be considered as in-
cluded in the proposition^ and as being the case when n=l.

PROP. V* THEOR.

If one magnitude be the same muUiple of another, y)hich a magnitude
taken from the first is of a magnitude taken from the other ; the re-
mainder is the same multiple of the remainder, that the whole is'of
the whole.

Let mA and mB be any equimultiples of the two magnitudes A and
B, of which A is greater than B ; mA— mB is the same multiple of
A— B that mA is of A, that is, mA— rmB=m (A— B).

Let D be the excess of A above B, then A— B=:D, and adding B
to both, A-=^D-f B. Therefore (1. 5.) mA=^mD4-mB ; take mB from
both, and mA-mB=mD ; but D=A— 6, therefore mA— mB=m(A
-B). Therefore, &c. a. £. D.

*-% *

114 ELEMENTS

PROP. VI. THEOR.

If from a multiple of a magnitudehy any number a multiple of the same
magnitude by a less number be taken auxty, the remainder wiU be the
same multiple of that magnitude that the difference of the numbers is
of unity •

Let mjL and nk. be multiples of the magnitude A, by the number?
m and n, and let m be greater than n ; mK — nk contains A as oil as
m — n contains unity, or mA — nA=(m— ii)A.

Let m — n=5^ ; then »i=n4-^« Therefore, (8. 6.) mA^nA+^A ;
take nk. from both, and mA — nA=igA. Therefore mA— nA contain^
A as ofl as there are units in q^ that is, in m— n^ormA — itA=(m— n)
A. Therefore, &c. Q, E. D.

Cor. When the difference of the two numbers is equal to unity,
or m — n=l, then mA— irA=A.

PROP. A. THEOR.

If four magnitudes be proportionals^ they are proportunuds also when^

taken inversely.

If A : B : : C : D, then also B : il : : D : C.

Let mA and mC be any equimultiples of A and C ; nB and nD any
equimultiples of B and D. Then, because A : B : : C : D, if mA
be less than nB, mC will be less than nD (def. 5. 5.), that is, if nB
be greater than mA, nD will be greater than mC. For the same rea-
son, if nB=mA, nD=mC, and if nBZ^A, nDZmC. But nB, nD
are any equimultiples of B andOD, and mA, mC any equimultiples of
A and C, therefore (def. 5. 5.), B : A : : D : C. Therefore, &c»
Q. E. D.

')
PROP. B. THEOR.

If the first be the same multiple of the secondy or the sttme part of it^
that the third is of the fourth ; the first is to the second as the third
to the fourth.

First, if mA, mB be equimultiples of the magnitudes A and B, mA
: A ; : mB : B.

Tike of mA and mB equimultiples by any number n ; and of A and
B equimultiples by any number jr; these will be nmA (3. 5.), pky
nmB (3. 5.), pB. Now, if nmA be greater than pA, nm is also
greater than p ; and if nm is greater than p, nmB is greater than jpB,
therefore, when nmA is greater than pA, nmB is greater than pB*
In.the same manner, if nmA==pA, nmB^pB, and if nmAZ.i'A, nmB
LpB* NoW; nm^i nmB are any equimultiples of mA and mB ; and

OP GEOMETRY. BOOK V.

115

pA, pB are any equimultiples of A and B, therefore tnA : A : : mB :
B (def.S. 6.)-

Next> Let C be the same part of A that D is of B ; then A is the
same multiple of C that B is of D, and therefore, as has been demon-
strated, A : C : : B : D and inversely (A. 5.) C: A : : D : B. There-
fore, dec. Q. E. D.

PROP. C. THEOR,

If the first he to the sec&nd as the third to the fourth ; and if the first he
a multiple or a part of the second^ the third is the same multiple or
the same part of the fourth*

Let A : B : : O : D, and Hrst, let A l>e a multiple of B, C is the
same multiple of D, that is, if A«=inB, C=mD.

Take of A and C equimultiples by any number as 2, viz. 2A and
2C ; aod of B and D, take equimultiples by the number 2m) viz. 2mBf
2ml) (3. 5.) ; then, because A=fnB, 2A=29nB ; and since A : B :*:
G : D, and since 2A=2mB, therefore 2C=2mD (def. 5. 5.), and C^
snD, that is, C contains D m times, or as often as A contains B.

I<{ext, Let A be a part of B, O is the same part of D. For, since
A : B : : C : D, inversely (A. 5.), B : A : : D : C. But A being a
part of B, B is a multiple of A ; and therefore, as is shewn above, D
is the same multiple of C, and therefore C is the same part of D
that A is of B. Therefore, 6cc. Q. E. D.

PROP. VIL THEOR-

Sgiki magnitudes haiee the same ratio to the same magnitude ; and the
same has the same ratio to egwd magnitudes.

Let A and B4>e equal magnitudes, and C anyother ; A : C : : B : Cl

Let mA, mB, be any equimultiples of A and B ; and nC any multi-
ple of, C.

Because A=B,iiiA=mB (Ax. 1. 5.) ; wherefore, if mA be greater
than nC, mB is greater than nC ; and if niA=ftC, mB=nC ; or, if
mAz.nC, mBZ.nC. But mA and mB are any equimultiples of A and
B, and nC is any multiple of C, therefore (def. 5. 5.) A : C : : B : C.

Again, if A=B, C : A : : C : B ; for, as has been proved, A :
C : : B : C, and inversely (A. 5.), C : A : : C : B. Therefore, dtc.
O. E. D.

PROP^Vm. THEOR.

Of unequal magnitudes^ the greater has a greater ratio to the same than
the less has ; and the same magnitude has a greater ratio to the less
than it has to.the greater,

Let A+B be a magnitude greater than A, and Q a third magnitudei

116

ELEMENTS

A+B has to C a greater ratio thaa AhxtB to C ; and C haa a greater
ratio to A than it has to A+B.

Let m be such a number that mA and mB are each of then greater
than C ; and let nC be the least multiple of G that exceeds mA+mB ;
then fiG— Cy that is (n-l) C (1. 5.) will be less than mA+«aB| or
mA+mB, that is, m(A+B) is greater than (n — 1)0. But heeause
itO is greater than mA+mB, and C less than mB, nC— C is greater
than fit A, or mA is less than »C — C, that is» than (it — 1)0. There*
fore the multiple of A+B by m exceeds the multiple of C by n — 1,
but the multiple of A hy m does not exceed the multiple of O by n— 1 ;
therefore A+B has a greater ratio to G than A has to G (def., 7. 5u)«

Again, because the multiple of C by n-* 1> exceeds the multiple of
A by m, but does not exceed the multiple of A+B by m, G has a great-
er ratio to A than it has to A+ B (def. 7. 5.) Therefore, &c. Q* iL D.

PROP. IX. THEOR.

Magnitudes which have the same ratio to the same magmiude are ejvol
to one another ; and those to which the same magnitude has the same
ratio are equal to one another • , ^

If A :G: :B:C, A=B.

For if not, let A be greater than B ; then because A is greater
than B, two numbers, m and it, may be found, as in the last propo*
sition, such that mA shall exceed nG, while mB dbes not exceed
nQ» But because A : G : : B : G ; and if mA exceed nG, mB must
also exceed nQ (def. 5. ^.) : and it is also shewn that mB does not
exceed nC, which is impossible. Therefore A is not greater than
B ; and in the same way it is demonstrated that B is not greater
thaa A ; therefore A is equal to B. ,

Next, let G : A : : G : B, A=^B. For by inverstoD (A. 5.) A \
G : : B : G ; and therefore, by the first case, A=B.

PROP. X. THEOR.

ThtA magnitude, which Juts a greater ratio than another has to the same
magnitude, is the greatest of the two : And that magnitude, to which
tJie same has a greater ratio than it has to another magmtude^is the
least of the two.

If the ratio of A to G be greater than that of B to G, A is greater
than B ,

Because A : G;7B : G, two numbers m and n may be found, such
that mA7nC, and mB^nG (def. 7. 6.), Therefore also mAvmB,
and A7B(Ax 4. 5.).'

Again, let G : B7G : A ; B^ A. For two numbers, m and n taajr
be found, such that mOynB, and mGz.nA (def. 7. 5.) Therefore,
since nB is less, and nA greater than the same magnitude mC, nB
^fiA, and therefore B^ A. Therefore, &c. * Q. E. D.

QF6EOMBTIIY. BOOK V.

117

HIOP. XI. THEOR. ,

BaUoi Am ure efuti tc^iheaame roAo are tfiui 1/9 am mmAerm

IfA:B::C:P; ftndahM>G:D::E:F; tben A : B : : £ : F.

Take mA, mC, mE^ any equimulUplea of A, C, and E ; a&d aB,
nD, nP, any equimnltiples of B, D, and F. Because A : B : : C : D^
if mA7itB,,iiiC-7nD /def. 5. 5.) ; but if wiCTnD, wETnF (def. 6.
5. ), because C : D : : E : F ; therefore if* mA 7 nB, foR 7 nF. In the
flame manner, if mA = nB, mE = nF \ and if mAZ^B, inE/.nF.
Now, mA, m£ are any equimultiples whatever of A and E; and nB|
nF any whatever of B and F ; therefore A : B : : £ : F (def. 5. 5c).
Therefore, d^c. Q* £. D»

PROP. XII. THEOR. .

If any number of magniiudeshe proportionaUi as one of the arUecedenU
is to its eonsequerUj so are aU the anteeedenis^ taken tt^iethery to aU
I&6 consequents* ^

If A : B s : G : D, and C ; D : : E : F ; then^so, A : B : : A+C
+E:B+D+F.

Take mA, tnCy mE any equimultiples of A, C, and E ; and siB» nD^
nF, any equimultiples of B, D, and^F. Then, hecause A : B : : C :
D, if mA7mB, nCynD (def. 5. 5.) ; and when i»C7nD, »iE7nF,
because C : D : : E : F. Therefore, if mA 7nB, mA+mC+mE 7nB
-hnD-f nF : In the same maimer, if mA=nB, mA+mC+mE=:«B4^
nD+nF; andif mAZ.nB,mA+mG+«EZnB+nD+nF. Now,mA
+wC+mE=m(A+C+E) (Cor. 1. 6.), so thatmA andmA+«C+
mE are any equimultiples of A, and of A +C + £• And for the same
reason nB^ and nB+nD+nF are any equimultiples of B, and of B+
D+ F ; therefore (def. 6. 5») A : p : : A + a+ E : B + D + F.
Therefore, tec. Q. £. D.

PROP. XIII. THEOR.

If the first have to the second the same ratio which the third has to ifte
fourth, but the^ihird to the fourth a greater ratio than the fifth has to
the sixth ; the first has also to the second a greater ratio than the fifth
has to the sixth*

If A : B : : C : D ; but C : D7E : F ; then also, A : B7E : P.

"Because C:D7E : F, there are two numbers m and n, such that
mC TnD, but mE Z. nF (def. 7. 5.). Now, if mC 7nD, mA 7nB, be.
cause A : B : : C ; D. Therefore mA7nB, and mE/.nF, where-
fore, A : B7£ : F (def. 7. 5.) Therefore, d(c. Q. £.D.

118 ELEMENTS

PROP. XIV. THEOR.

Iff ike first heme to the second the same ratio which the third has to the
fourth, and if the first ]be greater than the third, the second shall he
greater than the fourth ; ^ equal, equal ; and if less^ less.

IfA:B::C:D; then if AyC^ByD; ifA=C, B=D; and
ifAz.C, E/ID.

First, let A7C ; then A : BtC : B (8. 6.), but A : B : : C : D,
therefore C : D7C : B (13. 5.), and therefore B7D (10. 5.)

In the same manner, it is proved, that if A=C, B=D ; and if
AjLC, BZD. Therefore, &c. Q. E. D.

PROP. XV. THEOR.

magnitudes have the same ratio to one another which their equimiubipies

have.

If A and ^ be two magnitudes, and m any number, A : B : : mA :
-mB. •,

Because A : B : ; A : B (7. 5.) ; A : B : : A+A : B+B (12. 5.),
br A : B : : 2A : 2B. And in the same manner, since A : B : : 2A :
2B, A : B' : : A+2A : B+2B (12. 5.), or A : B : : 3A : 3B ; and so
on^ for all the equimultiples of A and B. Therefore^ &c. Q. £. D.

PROP. XVI. THEOR.

yfour magnitudes of the same kind be proportionals, they wiU also be

proportionals when taken qUenuxtely.

. If A :B : : C : D, then alternately, A : C : : B : D.
Take mA, mB any equimultiples of A and B, and nC, nD any equi-
multiples gf C and D. Then (15. 5.) A : B : : mA : mB; now A :
B: : C : D, therefore (11. 5.) C : D : : mA ; mB. But C : D : :
nC : hD (15. 5.) ; therefore mA : mB ; : nC : nD (11. 5.) : where-
fore if mA7nC, mB7nD (14. 5'.) ; if mA=nC,4nB=:nD, olrif mAZ.
nC, mB/^nJ); therefore (def. 5. 5.) A : C : : B : D. Therefore,
<Sz;c. Q. E. D.

PROP. XVII. THEOR.

If magnitudes, taken jointly , be proportionals, they wUl also bepropor*
iionals when taken iteparately ,\ th^ is, if the first, together with the
second, have to the second the same ratio which the third, iogether
with the fourth, has to tltt fourth, the first wUl have to the second the
same ratio which the third has to the fourth.

If A+B : B : : C+D ; D, then by division A : B : : C : D.

OF GEOMETRY, BOOK V. 119

Take mAand nB any multiples of A and B, by the numbers m and
n ; and first, letmAynB : to each of them add mB, then mA+mBy
mB+nB. But mA+OTB=«(A+B) (Cor. 1. 5.), and mB-f nB=
(m+n)B (2. Cor. 2. 5.), therefore «i(A+B) 7 {m+n)B.

And because A + B : B : : C+B : D, if »i(A + B)y{m+n) B,
«i(C+D)7(i»+n) D, or roC+wDTrnD+nD, that is, taking mD
from both, mCynD. Therefore, when tnA is greater than nB, mC
is greater than nD. In like manner it is demonstrated, that if mA=:
ftB, mC=nD, and if mAiinB, that mDz,nD ; therefore A : B : : C :
D (def. 5. 6.). Therefore, &;c. nQ. E. D-

PROP. XVIIL THEOR.

If magnttudfiSy taken^ separaieh/y he proportionals^ they toill also he pro*
partionals when taken jointly ^ that is, if the first be to the second 0$
the third to the fourth, the first and second together will be to the se»
cond as the third, and fourth together to the fourth*

If A : B : : C : D, then, by composition, A+B : B : : C+D : D.

Take iii(A-f'B)> and nB any multiples whatever of A+B. and B;
and first, let m be greater than n. Then, because A+B is also great-
er than B,ifi(A+B) 792B. For the same reason, m(C+D) ynD. In
this case, therefore, that is, when m^n, in(A+B) is greater than nB,.
and in(C+D) is greater than nD. And in the same manner it may
be proved, that when m=n, in(A+B) is greater than nB^ and m(C+
D) greater than »D.

Next, let mAn, or n7m, then m(A+B) may be greater than nB^
or may be equal to it, or may be less i first, let m(A+B) be greater
than nB ; then also, mA+mBT'nB ; take mB, which is less than nB,
from both, and mAynB — wB, or mA.'y {n—m)B (6. 5.). Bat if
nt A 7 (n — m) B, mC 7 (n— fn)D, because A : B : : C : D. Now, (n — m)
D=nD — mD (6. 5.), therefore mCynJ) — otD, and adding mD to
both, mC+mDTnD, that is (1. 6.), in(C+D)7nD. If, therefore,
»i(A+B) 7nB, in(C+D) 7nD.

In the same maner'it will be proved, that if m(A+B)=iiB^^(C+
D)=nD ; and if m(A+B)z.rtB, m(C+D)Znn ; therefore (def. 5.
6.), A+B : B : : C+D : D. Therefore, &;c. a. E. D»

PROP. XIX. THEOR.

If a whole magnitude he to a whole, as a magnitude taken from the first
is to a magnitude taken from the other ; the remainder will be to the
remamder as the whole to the whole.

If A : B : : C : D, ftnd if C be less than A, A— C : B-^D : : A : B

Because A : B : : C : D, alternately (16. 5.), A : C : : B : D ; and

therefore by division (17. 6.) A-^C : C : 2 B— D : D. Wherefore,

agaia alternately, A— C : B— D : : P : D ; but A : B : : C : D, there-

ISO BLEMENTS

fef^ (IL 5.) A-^ : B-^D : : A : B. Tberefbre^ 6^. Q. E. D.
Cm. a— O : B— D : : C ? D.

PROP. D. THEOR.

If four magntiude$ be proporUdnaU, Ae^itre dlseprajportionaUhyean*
vemotif Aat is, the/rH is to Us excess above the second^ as the third
to its egseess above thefcmttk.

If A : B : : C : D, by conversion, A : A— B : : C : C— D.

For, since A : B : : C : D, by division (17. 5.), A— B : B : : C— D
: D, and inversely (A. 6.), B : A — B : : D : C — D ; therefore, by com-
position (18. 6.), A : A- B : : C : C— D. Therefore^ &c Q. E. D.
. CojL In the same way, it may be proved that A : A+B : : C :
C+D.

PROP. XX. THEOR.

lfib»e k t&ree wagmbudesy and other ihree, %Meh ttdcen two amihoo^
hone the same roHa ; if the first he greater tkaxi the thirds thtfaurth
is greater than the siaSth ; if equals e^pud ; and ifless^ lus».

If there be three magnitudes, A, B, and C, and other three D, "E,

and F ; and if A : B : : D : E ; and also B : C j R — f^

: :J5 : F, then if A 7 C, D T F; if A = C, ' ' '

D^V; and if A 2iC,D2lF.

- D, E, F-

First, let A7C ; then A : B7C : B (8. 9.). But A : B : : D : E,
therefore also D : E7C : B (13. 6.). Now B : C : : E ; F, and invene*
ly (A. 5.), €:B : :F:E; and it has been shewn that D: E7€ : B,
Ifaereibre D : £ 7 F : E (13. 5.), and consequently D7F (10. 5.)

Next, let A=C ; then A : B : : C : B (7. 5.), but A : B : : D : E;
tlierefore, C : B : : D : E, butC : B : : F : E, therefore, D : E : : F :
E (VI. 6.), and D=F (9. 5.). Lastly, let AZC. Then C7A^
and because, as was already shewi^, C : B : : F : E, and B : A : : E :
D ; therefore, hy the first case, if 07 A, F-yD, that is, if AZC, D
Z F- Therrfore, ^ks, Q. E. D.

PROP. XXL THEOR.

tf there he three maghitttdes^ and other ihreey which have the same ra*
tio taken two and twoj but in a cross order ; ^ the first magniinde he
greater than the thirds the fourth is 'greater than the sixth ; if 6qualy
equal; and if less, less*

If there be three magnitudes, A, B, C, and other diree, D, E, and
P, such that A: B : : E : P,and B : C: : D : E; if A7C, D7F1
ifA=C,D=:P; andifAzCjP^F.

A, a

D, E,

C,
F.

OP GfiOJjETilY. ftOOK V. 121

iPiwt, let AtC. ' Tlr6n A : ByC : B (8. 5.),
feut A :^B : : E : P, therefore tJ : FyC t B
j(13. 5.). Now, B : G : : D : Ey and inversely,
C: : B : : E : D ; thetefore, E : F 7E « D (13. 5.), wherefol^, D 7F
(110.6.).

Next, let A=C, Then (?• 5.) A:B::C:B; but A: B :c E: P»
therefore^ C : 6 : : E : F (H. 5.) ; but B : C : : D : E, and inverse-
ly, C : B : : E : D, therefore (11. 6.), £ : F < : E : Di, ^nd, con8e>>
^uenlly, D=F (9. 5.).

Lastly, let A^C. Then CJ^A, and, as wa« already proved, C fi
B : : E : D ; and B : A : : F : E, therefore, by this first case, 8tiic«
C7 A, F^D, ^at is, DzF, Therefore, dec. Q. R Dv

PROP. VIIL THiEOR.

^ thete he any mmher of magnitudes, and as many others, td^tcA,
taken two and ttoo in order, have the same ratio ; the first win have
to the last ^ the first magnil'Ades, tJte same ratio tohich the first ^
the otJier has to the ieut.*

First, let there be three magnitudes. A, B, C, and other three, D^
Is, F, which, taken two and two, in order, have the satne ratio, viz»
A : B : : D : E> and B : C : : E : F ; then A : C ; : ID c F.

Take of A and D any equimultiples whatever, tnA, mD ; and of B
tind D any whatever, tiB, nF : and of O and F any whatever, qC^ 9F.
because A : B : : D : E, mA : nB : : mD : nE
(4. 5.); and for the same reason, rB : ^C : : nE :
i|^F« Therefore (20. 5. ) accordingas m A is great*
cr than qC, equal to it, ot less, inD is. greater
than ^F, equal to it or less ; but mA, mD are

A> B, C,
D, E, F,
fnA, nB, qC,
mP, nE, qF. \

any equimultiples of A and.D; dnd ^C, ^F are any equimultiples of
€ and F ; therefore (def. 5. 5.), A : C : s D : F.

Again, let there be four magnitudes, and other four which, taken
two and two in order, have the same ratio, viz. A : B c : B : F ; B :
C : : F : G ; O : D : : G : H, then A : D : J E : H.

For, since A, B, C are three magnitudes,
Und E, F, <jr other three, which, taken two
and two, have the same ratio, by the forego-
ing case, A : C : : £ : G. And because al-
so C : D : : G : H, by that same case, A : D : : E :.H. In the same
manner is the demonstration extended to any number of magnitudes*
Therefore, &c. Q. E* D.

• N. B. This propo«ilion is ii««iill^ cited by the words " en uequali," or " ex wquov**

122 ELEJffiNtS

PROP. XXIII. THEOR.

If there be any number of magniiudeis, and as many others^ whicttf
taken two and ttoOj in a cross order, have the same ratio ; the first
wUl have to the last of the first mag'miudes the same ratio whichJhe
first of the others has to the last.*

First, let there be three magnitudes, A, B, C, and other three, D,
E, and F, which, taken two and two in a cross order, have the same
ratio, viz. A : B : : E : F, and B ; C : : D : E, then A ^ C : : D : F.
•Take of A, B, and D, any equimultiples mA, mB> mD ; and of C,
E, F any equimultiples wC, nE, wF.

Because A : B : : E : F,and because also A : B : : mA : wiB (15. 5.),
and E : F : : nE : nF f therefore, mA : mB : : nE : nF (11. 5.) Again^
because B : C : : D : E, mB : nC : : ml>-: nE
{4. 5.) ; and it has been just shewn that mA :
mB: :nE:nF; therefore, if m A 7nC,mD"7nF
(21.5.); ifmA=nC,mD=nF; andifmAz.nC,
mD AnF. Now, mA and mD are any equimul-
tfples of A and D, and 7iG, nF any equimultiples of C and F ; there-
fore, A : C : : i) : F (def. 5. 5.).

Next, Let there be four magnitudes, A, B, C, and D, and other
fonr, E, F, G, and H, which, taken two and two in a cross order, have
the same ratio, viz. A:B::G.:H;B:C::
F : G, and C : D : : E r F, then A r D : : E :
■ H. For, since A, B, G, are three magnitudes,
and F, G, H, other three, which, taken two and

mA,

mB,

nC,,

mD,

nB,

«F.

A, B, C, D,
E, F, G, H.

two, in a cross order, have the aame ratio, by the first case, A : C : :
F : H. But C : D : : E : F, therefore, again, by the first case, A :
D : : £ : H. In the same manner may the demonstration be extend-
ed to any number of magnitudes. Therefore, <fcc. Q. E. J).

PROP. XXIV. THEOR.

If the first has to the second the same ratio which the third has to the
fourth ; and the fifth to the second, the same ratio which the sixth hasr
to the fourth; the first and fifth, together, shall have to the second, the
same ratio which the third and sixth together, have to the fourth.

Let A : B : ; C : D, and also E : B : : F : D, then A+E : B : : C+
F : D.

Because E : B : : F : D, by inversion, B : E : : D : F. But by hy-
pothesis, A : B : : C : D, therefore, ex aequali (22. 5.), A : E : : C :
F ; and by composition (18. 5.), A+E : E : : C+F : F. And again
by hypothesis, E : B : : F : D, therefore, ex aequali (22. 6.), A+E y
B : : C+F : D. Therefore, &c. Q. E. D.

♦ N. B* This'proposition is usually cited by the words " ex aequali in propordooe pertur-
ta :" or. '* ex. ».ntin i'ni7«>r«ilv''*

bata ;" or, '* ex. aecjuo inversely.

OF GEOMETRY. BOOK V. 128

PROP. E. THEOR.

if four magnitudes he proportioi/uilsj the sum of the first is to their
difference as the sum of the other two to their difference.

Let A : B : : C : D ; then if A7B, #

A+B : A-B : : C+D : C— D ; or if AZ B

A+B : B— A : : C+D : D~C.
For, if A7B, then because A : B : : C : D, by division (17. 6.),

A — B : B : : C — D : D, and by inversion (A. 6.),

B : A — B : : D : C — D. But, by composition (18. 5.),

A+B : B : : C+D : D, therefore, ex aequali (22. 5.),

A+B : A-B : : C+D : C— D.
In the same manner, if B7A, it is proved, that

A+B:B— A ::C+D:D--C. Therefore, &c.

Q. E. D. ;

PROP. F. THEOR.

lUuios which are compounded of equal ratios, are equal to one another*

Let the ratios of A to B, and of B to C, which compound the ratio
of A to C, he equal, each to each, to the ratios of D to E, and E to
F;^ which compound the ratio of D to F, A : C : : D : F.

For, first, if the ratio of A to B be equal to
that of D to S, and the ratio of B to C equal to
that of E to F, ex sequali (22. 5.), A : C : : D : F.

And next, if the ratio of A to B be equal to that of E to F, and
the ratio of B to C equal to that of D to E, ex aequali inversely (23.
6.), A : C : : D : F. In the same manner may the proposition bft
demonstrated, whatever be the number of ratios. Therefore, ^c.
ij. E. D,

ELEMENTS

GEOMETRY.

BOOK VL

DEFINITK)NS.

^iHiLAir rectilinear figures
are those which have their
several angles equal, each to*
each, and the sides about the
eqjatd angles pisoportionals.

Two sidias of one figure are said to be reeiprocally proportional to*
two sides of another, when one of the sides of the first is to one oT
the sides of the second-, as the remaining side of the second is to»
the remaining side of the first..

IlL
A. straight line is said to be eut in extreme andmean ratio-, when the-
whole is to the greater segment, ad>the greater segment is t<^the ks8.u

lY.

The altitude of any figure is the straight line
drawn, from its vertex perpendicular to the:
base^

PROP. L THEOR.

Triangles and parallelogramSi of the same altitude^ are one to another-

as their hoses*

Let the triangles ABC, ACD, and the parallelograms EC, CFhave-
the same altitude, viz. the perpendicular drawn from the point A
to BD I Then, as the base BC, is to the base CD, so is the trianglo^

ELEMENTS, &;c. 125

AB€ to the triangle ACD, and the parallelogram EC to the paral-
lelogram CF.

Pjroduce BD both ways to the points H, L, and take any number of
straight lines BG,GH, each equal to the base BC ; and DK, KL, any
number of them, each equal to the base CD ; and join AG, AH, AK,
AL. Then, because CB, BG, GH are all equal, the triangles AHO,
AGB, ABC are all equal (38. 1.) ; Therefore, whatever multiple the
base HC is of the base BC, the same multiple is the triangle AHC
of the triangle ABC. For the same reason;^ whatever the base
LC is of the base CD, the tt a "C^

same multiple is the tri* Ji A. Ji

angle ALC of the triangle
ADC. But if the base
HC be equal to the base
CL, the triangleAHC is
also equal to the triangle
ALC (38. 1.) ; and if
the base HC be greater --. -, ^ p,
ihan the base CL, likewise H w Ji v/
the triangle AHC is great-
er than the triangle ALC ; and if less, less. Therefore, since there
are four magnitudes^ viz. the two bases BC, CD, and the two trian-
gles ABC, ACD ; and of the base BC and the triangle ABC, the
iirst and third, any equimultiples whatever have been taken, viz. the
base HC, and the triangle AHC ; and of the base CD and triangle
ACD, the second and fourth, have been taken any equimultiples
whatever, viz. the base CL and triangle ALC ; and since it has been
shewn, that if the base HC be greater than the base CL, the triangle
AHC is greater .than the triangle ALC ; and if equal, equal ; and if
less, less ; Therefore (def. 5. 5.), as the base BG is to the base CD^
so is the triangle ABC to the triangle ACD.

And because the parallelogram CE is double of the triangle ABC
(41. 1.), and the parallelogram CF doublc-of the triangle ACD, and
because magnitudes have the same ratio which their equimultiples
have (15. 5) ; as the triangle ABC is to the triangle ACD, so is the
parallelogram EC to the parallelogram CF. And because it has been
shewn^ that, as the base BC is to the base CD, so is the triangle ABC
to the triangle ACD ; and as the triangle ABC to the triangle ACD,
so is the parallelogram EC to the parallelogram CF ; therefore, as
the base BC is to the base CD, so is (11. 6.) the parallelogram EC
to the parallelogram CF. Wherefore triangles, '&c. Q. E. D.

Cor. From this it is plain, that triangles and parallelograms that
have equal altitudes, are to one another as their bases.

Let the figures be placed so as to have their bases in the same
straight line ; and having drawn perpendiculars from the vertices of
the triangles to the bases, the straight line which joins the vertices is
parallel to that in which their bases are (33. 1.), because the per-
pendiculars are both ec^uQl and parallel to one another. Then, if the

126

ELEMENTS

flame construction be made as in the proposition, the demonstration
vill be the same,

PROP. II. THEOIL

If a straight litie be drawn parallel to one of the sides of a triangle^ U
wiU cut the other sidefiy or the other sides produced, proportionally :
And if the sides, or the sides produced, he cut proportionally, the,
straight line which joins the points of section will be parallel to the
remaining side of the triangle.

Let BE be drawn parallel to BC, one of the sides of the triangle
ABC : BD is to DA as CE to £A.

Join BE, CD ; then the triangle BDE is equal to the triangle CDE
(37. 1.), because they are on the same base DE and between the
flame parallels DE, BC : but ADE is another triangle, and equal
magnitudes have, to the same, the same ratio (7. 5.) ; therefore, as
the triangle BDE to the triangle ADE, so is the triangle CDE to the
triangle ADE ; but as the triangle BDE to the triangle ADE, so is
(1. 6.) BD to DA, because, having the skme altitude, viz. the perpen-
dicular drawn from the point E to AB, they are to one another as
their bases ; and for the same reason, as the triangle CDE to the
triangle ADE, so is CEJ to £ A. Therefore, as 9D to DA, so is CE
toEA(I1.5.),

^ntf let the aides AS, AC of the triangle ABC, or these side%

O D

E B

produced, be cut proportionally in the points D, E, that is, so that
BD be to DA, as CE to EA, and join DE ;. DE is parallel to BC.

The same construction being made, because as BD to JDA, so is C£
to EA ; and as BD to DA, so is the triangle B1)E to the triangle ADE
(I. 6.) : and as CE to EA, so is the triangle CDE to the triangle
ADE ; therefore the triangle BDE, is to the triangle ADE, as the tri-
angle CDE to the triangle ADE ; that is, the triangles BDE, CDE
have the same ratio to the triangle ADE ; and therefore (9. 5.) the
triangle BDE is equal to the triangle CDE : And they are on the same
bftse DE } but equal triangles on the same base ajo between tha

5f GEOMETRY. BOOK Vt. ISff

Satne parallels (39. 1.) ; therefore DE is parallel to BC. Where-
fore) if a straight line, &c. Q. £. D«

PROP. lU. THEOR.

tj the angle of a triangle he bisected hy a straight line whicTi also aSUs >
the base ; the segments of the base shall have the $ame ratio winch the
other sides of ike triangle have to one another ; And if the segments
of the base have the same ratio which the other sides of the triangle
have to one another^ the straight line drawn from the vertex t6the
point of section, bisects the vertical angle.

Let the angle BAG, of any triangle ABC, he divided into two equal'
angles, by the straight line AD ; ^D is to DC as BA to AC.

Through the point C draw CE parallel (31. 1.) to DA, and leiBA
produced meet CE in E. Because the straight line AC meets the
parallels AD, EC, the angle ACE is equal to tha alternate angle
CAD (29. 1.) : But CAD, by the hypothesis, is equal to the angle
BAD ; wherefore BAD is equal to the angle ACE. Again, because:
the straight line BAE meets the
parallels AD, EC, the exterior angle
BAD is equal to the interior and
opposite angle AEC ; But the angle
ACE has been proved equal to the
angle BAP ; therefore also ACE is
equal to the angle AEC, and conse-
quently the side AE is equal to the
side (6. 1.) AC. And because AD
is drawn parallel to one of the sides

of the triangle BCE,'viz. to EC,,^^ '■ =jJr ^

BD is to DC, as BA to AE (2. 6.) ; -tS U U

but AE is equal to AC ;, therefore, as BD to DC, so is BA to AC

(7. 5.).

Next, let BD be to DC, as BA to AC, and join AD ; Ihe angle
BAC is divided iato two equal angles, by the straight line AD.

The same construction being made ; because, as BD to DC, so is
B A to AC ; and as BD to DC, so - ' xp

is BA to AE (2. 6.), because AD ^^

is parallel to EC : therefore AB
is to AC, as AB to AE (11. 5.) :
Consequently AC is equal to AE
(9. 5.), and the angle AEC is
therefore equal to the angle ACE
(5. 1.) But the angle AEC is
equal to the exterior and oppo-
site angle BAD; and the angle

ACE is equal to the alternate an* B DC-

'gle CAD (29. 1.) : Wherefore also the angle BAD is equal to the an.

m ELEMENTS

gle CAD : Therefore the angle BAC is cut into two equal angtcts hf
the straight line AD. Therefore, if the angle^ &c. Q, E. D.

PROP. A. THEOit.

if the ezteriat angle qf a triangle he hUected bp a draighl line which
also euU.the hatie produced; the segmfints bettoeen the bisecting lin6
and the extremities of the base have (he same ratio which the other
sides of the triangles have to one another ; And if the segments of
the base produced have the same ratio which the other sides of the
triangle have^ the straight IvnCy drawn from the vertex to the point of
section^ bisects the exterior angle of the triangle.

Let the exterior angle CAE, of any triangle ABC, be bisected by
the straight line AD which meets the base produced in D ; BD is to
DC, as BA to AC.

Through C draw CF parallel to AD (31. 1.): and because the
straight line AC meets the parallels AD, FC, the angle ACF is equal
' to the alternate angle CAD (29. 1.) : But CAD is;equal to the angle
t)AE (Hyp») : therefore also DAE is equal to the angle ACF. Again^
because the straight line FAE meets the parallels AD, FC, the exte->
rior angle DAE is equal to the interior and opposite angle CFA ; But
the angle ACF has been prov-
ed to be equal to the angle
DAE ; therefore also the angle
ACF is equal to the angle CFA,
and consequently the side AF
is equal to the side AC (6. L) ;
and, because AD is parallel to
FC, a side of the triangle BCF,
BD is to DC, as BA to AF (2. jf
6.) ; but AF is equal to AC ; ""
therefore as BD is to DC, so is BA to AC.

Now Jet BD be to DC, as BA to AC, and join AD ; the angle CAD
is equal to the angle DAE.

The same construction being made, because BD is to DC, as BA
to AC ; and also BD to DC, BA to AF (2. 6.) ; therefore BA is to
AC, as BA to AF (IL 5.) wherefore AC is equal to AF (9. 5.), and
the angle AFC equal (5. 1.) to the angle ACF : but the angle AFC
is equal to the exterior angle E AD, and the angle «ACF to the alter-»
nate angle CAD; therefore also EAD is equal to the angle CAD<
Wherefore, if the exterior, &c. Q. E. D.

PROP. IV. THEOR-

The sides about tJie equal angles of equiangular triangles are propdf*
tUmcds ; and those which are opposite to l/te equal angles are homolo*
gous sides, that is, are the antecedents or consequents cfthe ratios.

Let ABC, DCE, be equiangular triangles, having the angle ABC

OP GEOMETRY. BOOK Yl.

S»

^t|aal to the angle DCE, and the angle ACB to the angle DEC, and
^onsequetitly (32. 1.) the angle BAG equal to the angle ODE. The
^ides about the equal angles of the triangles ABC^ DCE are^ propor*
tionals * and those are the homologous sides which are opposite te
the equal angles.

Let the triangle DCE be placed, so that its side CE may be conti-
guous to BC) and in the same straight line with it : And because the
angles ABC*, ACB are together less than two right angles (17^ 1 .\ ABC
and.DEC, which is equal to ACB, are al- J*)
«K> less than two right angles : wherefore
BA, ED produced shall meet (Cor. 29. 1.) ^
let them be produced and meet in the
point F ; and because the angle ABO is
*equal to the angle DCE> BF is parallel
f28. 1.) to CD. Again, because the an<-
gle ACB is equal to the angle DEO, AC is
parallel to FE (28. 1.) : Therefore FACD
is a parallelogram,; and consequently AF B
is equal to CD, and AC to FD (34. 1.) :
And because AC is parallel to FE^ one of the sides of the triangle
FBE, BA : AF : : BO : CE (2. 6.) ; but AF is equal to CD : there-
fore (7. 5.) BA : CD : : BO : OE ; and alternately, BA : BO : : DC
: CE (16. 5.) : Again, because CD is parallel to BF, BO : CE : :
FD : DE (2. 6.; ; but FD is equal to AC 5 therefore BO : CE : : AP
: DE ; aad alternately, BC :'CA $ : OE : ED. Therefore, because
it has been proved that AB : BO : : DC : CE ; and BC : CA : : CS
: ED, ex sequali, BA ^ AC : : CD : DE. Therefore the sides* dec.
Q, p. D.

PROP. y. THEOR.

if the sides of two triangles, dboui e^uh of their angles, he proportioik^
alSj the triangles shall be equiangular, and have their eqtud angles
opposite to the homologous sides*

Let the triangles ABC, DEF have their sides ptoportionals* so that
AB is to BO, as DE to EF ; and BC to OA, as £F to PD^ and ,
consequently, ex asquali, BA to.AC, as ED to DF ; the triangle ABC
IS equiangular to the triangle DEF, and their equal angles are op-
posite to the homologous sides, viz. the angle AMfi being equal to
eke angle DEF, and BOA to EFD, and also JBAC to EDF.

isa

ELEMENTS

At the points E, F, in the
fllraight line EF, make (23. 1.)
the angle FEG equal to the an-
gle ABC».and the angle EFG
equal to BCA, wherefore the re-
maining angle BAG is equal to
the remaining angle EGF (32.
1.), and the triangle ABCis there,
fore equiangular to the triangle
6£F; and consequently they
have their sides opposite to the
equal angles proportionals (4. 6.). Wherefore,

AB : BC : : GE : EF ; but by supposition,

AB : BC : : DE : EF, therefore,

DE : EF : : GE : EF. Therefore (11. 5.) DE and GE
have the same ratio to EF, and consequently are equal (9. 5.). Fbr
the same reajson, DF is equal to FG : And because, in the triangles
DEF, GEF, DE is equal to EG, and EF common, and also the base
DF equal to the base GF; therefore the angle DEF is equal (8. 1.)
to the angle GEF, and the other angles to the other angles, which are
subtended by the equal sides (4. 1.) Wherefore the angle DFE is
equal to the angle GrFE, and EDF to EGF : and because the angle
DEF is equal to the angle GEF, and GEF to the angle ABO ; there-
fore the angle ABC is equal to the angle DEF : For the same rea-
flon, the angle AGB is equal to the angle DFE, and the angle at A-
to the angle at D. Therefore the triangle ABC is equiangular to
die triangle DEF* Wherefore, if the sides, &c. Q. £. D.

PROP. VI. THEOR.

If two triangles have one angle of the one equal to one angle of the other,
and the sides about the equal angles proportionals , the triangles shaU
he equiangular y and shaU have those angles equal Khich are opposite
to' the homologous sides.

Let the triangles ABC, DEF have the angle B AC in the one equal
to the angle EDF in the other, and the sides about those angles pro-

?ortionals ; that is, BA to AC, as ED to D^ ; the triangles ABC,
)EF are equiangular, and have the angle ABC equal to the angle
DEF, and ACB to DFE.

At the points D, F, is j^
the straight line X)F, make
(23. 1.) the angleTOG dqual
to either of the angles BAC^
EDF ; and the angle DFG
equal to the angle ACB;
wherefore the remaining
angle at B is equal to the
remaining one at G (32. 1.),
and consequently the trian-

OPOEOMETRT. BOOK VI. 181

gle ABC! is eqaiangutair to the triamlo D6F ; and therefore
BA : AC : : GD (4. 6.) : DP. But by hypothesis,
BA : AC : ; ED : DF ; and therefore

ED : DP :: GD : (1 1. 5.) DF ; wherefore ED is equal (9. 5.)
to DG : and DP is common to the two triangles EDF^ GDF ; there-
fore the two sides ED, DF are equal to the two sides GD, DF ; but
the angle EDF is also equal to the angle GDF ; wherefore the base
EF is equal to the base FG (4. 1.), and the triangle EDF to the tri-
angle GDF, and the remaining angles to the remaining angles, each
to each, which are subtended by the equal sides.: Therefore the
angle DFG is e^tfal to the angle DFE, and the angle at G to the angle
at E : But the angle DFG is equal to the angle ACB ; therefore the
angle ACB is equal the angle DFE, and the angle BAC is equal to the
angle EDF (Hyp.) ; wherefore also the remaining angle at B is equal
to the remaining angle at E. There/ore the triangle ABC is equianl
gular tothe triangle DEF. Wheretbre, if twb triangles, dsc. Q.E. D»

PROP, VIL THEOR.

If two triangles have one angle of the one equal to one angle of the other y
and the sides about two other angles proportioneds, then, if each of the
remaining angles he either less, or not less, than a right angle, the tri-
angles shall be equiangular, and have those angles equal about which

. t?ie sides are proportionals*

Let the two triangles ABC, DEF have one angle in the one equal to
one angle in the other, viz. the angle BAC to the angle EDF>and the
sides about two other angles ABC, DEF proportionals, so that AB is
to EC, as DE to EF ; and, in the first case, let each of the remaining
angles at C, F, be less than a right angle. The triangle AfeC is equi-
angular to the triangle DEF, that is, the angle ABC is equal to the an.
gle DEF, and the remaining angle at C to the remaining angle at F»

For, if the angles ABC, DEF be not equal, one of them is greater
than the other : Let ABC be the greater, and at the point B, in the
straight line AB, make the angle
ABG equal to the angle (23. 1.)
DEF : and because the angle at A
is equal to the angle at D, and the
angle ABG to the angle DEF ;
the remaining angle AGB is
equal (32. 1.) to the remaining
angle DFE ; Therefbre the tri-
angle ABG is equiangular to the
triangle DEF ;

wherefore (4. 6.), AB : BG : : DE : EF ; but,
- by hypothesis, DE : EF : : AB : BC^ .
therefore^ AB : BC ; ; AB ; BG (11. 6.)

183

ELEMENTS

and because AB has the same ratio to each of the lines BC, BG ; BC
18 equal (9. 5.) to B6, and therefore the angle BGC is equal to the an-
gle BCG (6. 1.) ; But the angle BG6 is, by hypothesis, less than a
right angle ; therefore also the angle BGC is less than a right angle,
and the adjacent angle AGB must be greater than a right angle (13;
1.). But it was prored that the angle AGB is equal to the angle at F ;
therefore the angle at F is greater than a right angle : But by the
hypothesis, it is less than a right angle ; which is absurd. Therefore
the angles ABC, !DEF are not unequal, that is, they are equal : And
the angle at A is equal to the angle at D ; wherefore the remaining
angle at C is equal to the remaining angle at F ; ^erefbre the tri-
angle ABC is equiangular to the triangle PEF.

Next, let each of the angles at C, F be not less than a right angle ;
the triangle A BC is also, in this case, equiangular to the' triangle DEP.

The aame construction b^in^^ made, it .may be proved, in like ihaB*>
ner, that fiC is equal to BG,
and the aiigle at C equal to
the angle BGC : But Uie an-
gle at C is not less than a right
angle ; therefore the angle
BGC is not less than a right
> angle :* . Wherefore, two an-
gles of the triangle BGC are
together not less than two right angles, which is impossible (17, L) ;
and therefore the triangle ABC may be proved to be equiangular Uk
the triangle DEF, as in the first '<;a8e.

PROP. Vni. THEOR,

Jh aright angled triangle if a perpendicular he drawn from the rigM
angle to the base ; the triangles on each side of it are simitar to tHe
whole triangle^ and to one another.

Let ABC be a right angled triangle, having the right angle BAC ;
and from the point A let AD be drawn perpendicular to (he base BO:
the triangles ABD, ADC are similar to the whole triangle ABC, an^
to one another.

Because the angle BAC is equal to the angle ADB, eaeh of thena
being a right angle, and the angle J^

at B common to the two trianglea
ABC, ABD : the remaining ansle
ACB is equal to tbe remaining an-
gle BAD (82. 1.) : therefore the
triangle ABC is equiangular to th^
triangle ABD^ and the sides about ]g
their equal angles are proportion,
als (4. 6.) J wherefore the triangles are similar (def. 1. 6.). In lik«
manner; it may be demonstrated; that the triangle ADC » equiaa^

OP GEOMETRY. BOOK VI.

18S

gular and similar to the triangle ABC : and the triangles ABD, ADC»
being each equiangular and similar to ABC, and equiangular and si.
milar to one another. Therefore, in a right angled, &c. Q. E. D.

CoR. From this it is manifest, that the perpendicular, drawn from
the right angle of ,a right angled triangle, to the base, is a mean pro-
portional^ between the segments of the base ; and also that each of
the sides is a mean proportional between the base, and its segment
adjacent to that side. For in the triangles BDA, ADC,

' BD : DA : : DA : DC (4. 6.) ; and in the
triangles ABC, BDA, BC : BA : : BA : BD (4* 6.) ; and in th»
triangles ABC, ACD, BC : CA : : CA : CD (4. 6.).

PROP. IX. PROB.

From a given straight line to cut of any part requiredy that is, apian
which shall be contained in it a given number of times.

Let AB be the. given straight line ; it is required to cut off from
AB, a part which shall be contained in it a given number of times.

From the point A draw a straight line AC making any angle with
AB ; and in AC take any point D, and take AC
such that it shall contain AD, as ofl as AB is to
contain the part, which is to be cutofffcomit;
join BC, and draw D£ parallel to it : then AE
is the part required to be cut off.
' Because ED is parallel to one of the sidies
of the triangle ABC, viz. to BC, CD : DA : :
DE : EA (2. 6.) ; and by composition (18. 5.),
CA : AD : : BA : AE : But CA is a multiple
of AD ; therefore (C. 5.) BA is the same
multiple of AE, or contains AE the same num-
ber of times that AC contains AD ; and there-
fore, whatever part AD is of AC, AE is the same of AB ; wherefore,
from the straight line AB the part required is cut off. Which was
to be done.

PROP. X. PROB.

To divide a given straight line similarly to a given divided straight line\
that is, into parts that shall have the same ratios to one another which
the parts of the divided given straiglU line have.

Let AB be the straight line given to be divided, and AC the divided
line, it is required to divide AB similarly to AG.

Let AC be divided in the points D, E ; and let AB, AC be placed
so as to contain any angle, and join BC, and through the points D, E,
drawn (3L 1.) DF, EG, parallel to BC ; and through D draw DHK,
parallel to AB ; dierefore each of the figures FH, HB, is a parallelo-

w ,

184

ELEMIINTS

fram : wherefore DH id equal (34. 1.) to /^
G, and HK to GD : aad because HE is '^
parallel to KC, one of the sides of the tri-
angle DKC, CE : ED : : (2. 6.) KH : HD ;
But KH=BG, and HD=GF ; therefore J*
C£ : ED : : BG t GF ; Again, because
FD is parallel to EG, one of the sides of ri
the triangle AGE, ED : DA : : GF : PA ; ^
But it has been proved that CE : ED : :

B(jr : OF ; therefore the given straight line ^-d
AB is ditided similarly to AG. Which was -^
to be done. '

PROP. XI. PROB.

To find a third proportional to tvx) given straight lines.

Let AB, AC be the two given straight lines, and let them be placed
so as to contain any angle ; it is requir- J^ /
ed to find a third proportional to AB, AC.

Produce AB, AC to the points D, E ;
and make BD equal to AC ; and having «^
joined BC, through D draw DE parallel ^
to it (31. 1.).

Because BC is parallel to DE, a side
of the triangle ADE, AB : (2. 6.) BD : :
AC : CE ; but BE=AC : therefore AB :
AC : : AC : CE. Wherefore to the two
given straight lines AB, AC a third pro-
portional, CE is found. Whick wais to be done.

PROP. XIL PROB.

Tofiind a fourth proportional to three given straight lines.

Let A, B, C be the three given straight lines ; it is required to fin d
a fourth proportional to A, B, C.

Take two straight lines DE, DF, cpntaining.any angle EDF ; and
upon these make DG equal to A, GE equal to B, and DIf equal to C ;
and having joined GH, draw EF parallel (31. 1.) to it through the

OF GEOMFTRY. BOOK YI. 185

pxnni E. And because 6H is paraHol to EF, one of
triangle DEF, DG ; GE : : DH : HF (2. 6.) ; but 1)G=:A, GE=B,
and DH=bC ; and therefore A : B : : C : HF. Wherefore to the
three given straight lines, A« B, C> a fourth propertional HF is founds
Which was to be done.

PROP. XIII. PROB.

Toj^nd a mean proportional heHoeen Uoo given straight Imes.

Let AB, 6C be the two given straight lines ; it is required to find a
knean proportional between them.

Place AB, BC in a straight line, and upon AC desctibe the semiciir-
cle ADC, and from the point B
(11. 1.) draw BD at right angles
to AC, and join AD, DC.

Be'cause the angle ADC in a se- ,
micircle is a right angle (31. 3.).
and because in the right -angled
triangle ADC, DB is drawn from
the right angle, perpendicular to
the base, DB is a mean propor-
tional between AB, BC, the segments of the base (Cor. 8. 6.) ; there-
fore between the two given straight lines AB, BC, a mean proportion-
al DB is founds Which was to be done.

PROP. XIV. PROB.

•

Equal pardUdograms which have one angle of the one equal to one^an^e
of the other J have their sides about the equal angles reciprocally prO'
portiondl : Andparallelograifns which have one angle of the one equal
to one angle of the other y and their sides about the equal angles red*
procallp proportionaly are equal to one another.

Let AB, BC be equal parallel- j^
ograms, which have the angles
at B equal, and let the sides DB,
BE be placed in the same straight
line ; wherefore also FB, BG
are in one straight line (14. 1.) ;
the sides of the parallelograms
AB, BC, about the equal angles
are reciprocally proportional ;
that is, DB is to BE, as GB to
BF.

Complete the parallelogram FE ; and because the parallelograms
AB, BC are equal, and FE is another parallelogram,

AB : FE 2 : BC : FE (7. 5.) 5

IM ELEMENTS

but because the parallelograms AB, FE have the same altitude^

AB : FE : : DB : BE (L 6.) also,
BC : FE : : GB : BF (L 6.) ; therefore
DB : BE : : GB : BF (U. 6.). Wherefore, the
sides.of the parallelograms AB, BC about their equal angles are reci*
procally proportional.

But, let the sides about the equal angles be reciprocally propor-
tional, viz. as DB to BE, so GB to BF ; the parallelogram AB is
to the parallelogram BC.,

Because, DB : BE : : GB : l^F, and DB : BE : : AB : FE, and
GB : BF : : BC : EF, therefore, AB : FE : : BC : FE (11. 5.) :
wherefore the parallelogram AB is ^qual (9. 5.) to the parallelogram
BC. Therefore equal parallelograms, 6ic* Q. E. D.

PROP. XV. THEOR.

Equci triangles tDhich have one angle of the one equal to one angle of the
other have their sides about the equal angles reciprocally proportion'
al : And triangles which have one angle in the one equal to one angle
in the other, and their sides about the equal angles- reciprocally pro*
portional, are equal t& one another »

Let A'BC, AD£ be equal
triangles, which have the an-

fie BAG equal to the angle
^AE : the sides about the
equal angles^of the triangles
are reciprocally proportion-^ /f
al ; that is, CA is to AD, as ^
£A to AB.

Let the triangles be plac-
ed so that their sides CA^ ^
AD be in one straight line ; *
wherefore also EA and AB
are in one straight line (14. B *

I.) ; join BD. Because the triangle ABC is equal to the triangle
ADE, and ABD is another triangle ; therefore, triangle CAB : trian-
gle BAD : : triangle EAD : triangle BAD ; but CAB : BAD : : CA :
AD, and EAD : BAD : : EA : AB ; therefore CA : AD : : EA : AB
(11. 5.), wherefore the sides of the triangles ABC, ADE about the
equal angles are reciprocally proportional.

But let the bides of the triangles ABC, ADE, about the equal arr^
gles be reciprocally proportional, viz. CA to AD, as EA to AB ; the
triangle ABC is equal to the triangle ADE.

Having joined BD as before ; because CA : AD : : E A : AB ; and
since CA ; AD : : triangle ABC : trianrie BAD (1. 6.) j and also
EA :;AB : : triangle EAD : triangle BAD (11.5.); therefore, triaa-
gle ABC : triangle BAD : : triangle EAP : triangle BAD \ that ia.

OP GEOiftEtRY. BOOK VI.

list

^e triangles ABC, E AD have the same ratio to the triangle BAD ;
wherefore the triangle ABC is equal (9. 5.) to the triangle BAD.
Therefore equal triangles, &c. Q. E. D. *

PROP. XVI. THEO&.

if four straight lines he proportionaJsy the rectangle contained iy the
•extremes is equal to the rectangle Contained by the means ; And if the
rectangle contained by the extremes be equal to the rectangle contain^
ed by the means, the four straight lines are proportionals.

Let the four straight lines, AB, CD, £, F, he propottionials, viz. oa
AB to CD, so E to F ; the rectangle contained by AB, F is equal to
the rectangle contained by CD, E. ^ i««;$iK^|^

From the points A, C draw (11. 1.) AG, CH at right angles to AB>
CD ; ^nd make AG equal to F, and CH equal to E, and complete the
parallelograms BG, DH. Because AB : CD : : E : F ; and sii^ce
E=CH, and F=AG, AB : CD (7. 5.) : : CH : AG ; therefore the
sides of the parallelograms BG, DH about the equal angles are reci.
procally proportional; but parallelograms which have their sides about
equal angles reciprocally proportional, are equal to one another (14.
6.) ; therefore the parallelogram BG is equal to the parallelogram
DH ; and the parallelogram DG is "fg-
coataiaed by the straight lines AB,
F/; because AG is equal to P ; and '^-
the parallelogram DH is contained
by CD and E, because CH is equal
to E : therefore the rectangle con-
tained by the straight' lines AB, F
is equal to that which is contained
by CD and E.

And if the rectangle contained

by the straight lines AB, F be equal to that which is contained by CDy
E ; these four lines are proportionals, viz. AB is to CD, as E to F.
The same construction being made, because the rectangle contain^
ed by the straight lines AB, F is equal to that which is contained by
CD, B, and the rectangle BG is contained by AB, F, because AG is
equal to F ; and the rectangle DH, by CD, E, because CH is equal
to E ; therefore the parallelogram BG is equal to the parallelogram
DH, and they are equiangular : but the sides about the equal angles
of equal parallelograms are reciprocally proportional (14. 6.) t
wherefore AB : CD : : CH : AG ; but CH5=E, and AG=F, tiere.
fore AB : CD : : E : F. Wherefore, if four^ &c. Q. E* D.

138 ELEMENTS

PROP. XVIL THEOR.

If three straight lines he proportionals^ the rectangle contained hy the
extremes is equal to the square of the mean : And if the rectangle
contained by the extremes be equal to the square of the mean, the three
straight lines are proportionals.

Let the three straight lines, A, B, C be proportionals, viz. as A to
B, so B to G ; the rectangle contained by A, C is equal to the square
of B.

Take D equal to B : and because as A to B, so B to C, and that B
is equal to D ; A is (7. 5.) to B, as D to C : but if four straight lines
be proportionals, the rectangle contained by the extremes is equal to

that which is contained by the means (16. 6.) ; .

therefore the rectangle A.C = the rectangle j^ _

B.D ; but the rectangle B.D is equal to the j-.

square of B, because B=D ; therefore the rect- p _^____^

angle A.C is equal to the square of B.

And if the rectangle contained by A, C be equal to the square of B ;
A : B : : B : C.

The same construction being made, because the rectangle contain-
ed by A, G is equal to the square of B, and the square of B is equal:
to the rectangle contained by B, D, because B is equal to D ; there-
fore the rectangle contained by A, C is equal to that contained by B,
D ; but if the rectangle contained by the extremes be equal to that
contained by the means, the four straight lines are proportionals (16.
6.) : therefore A :B : : D : C, but Bs^D ; wherefore A : Br: B : G ;
Therefore, if three straight lines, &c. Q. £. D.

PROP. XVIII. PROB.

Upon a given straight line to describe a rectttineaH figwrt similar^ and
similarly situated to a given rectilineal figure.

Let AB be the given straight Hne, and CDEF the given rectilineal
figure of four sides ; .it is required upon the given straight line AB Xo
describe a rectilineal figure similar, and similarly situated to CDEF.

Join DF, and at the points A, B in the straight line AB, make (23.
1») the angle BAG equal to the angle at C, aqd the angle ABG equal
to the angle CDF; therefore the remaining angle CFD is equal to the
remaining angle AGB (32. 1 .) : wherefore the triangle FCD is equi.
angular to the triangle GAB: Again, at the points G, B in the straight
line GB make (23. 1.) the angle BGH equal to the angle DFE, and
the angle GBH equal to FDE ; therefore the remaining angle FED is
equal to the remuning angle GHB, and the triangle FDE equiangular
to the triungle GBH : then, because the angle AGB is equal to the
angle CFD, BGH to DF£, the whole angle AGH is equal to the

OF GEOMETRY. BOOK VI.

18d

L F

whole CFB : for the saine reason, the angle ABH is equal to the
angle CDE ;* also the angle at A is equal to the angle at C, and the
angle GHB to FED ; Therefore the rectilineal figure ABHG is equi-
angular to CDCF : but likewise these figures have thrir sidos about
the equal angles proportionals : for the triangles GAB, FCD being
•equiangular,

BA : AG : : DC : CF (4. 6.) ; for the same reason,
AG : GB : : CF : FD ; and because of the equi-
angular triangles BGH, DFE, GB : GH : : FD : FE ; therefore,

ex ffiquali (22. §.) AG : GH : : CF : FE.
In the same manner, it may be proved, that

AB : BH : : CD : DE. Also (4. 6.),
GH : HB : : FE : ED. Wherefore, because the rec-
tilineal figures ABHG, CDEF are equiangular, and have their sides
about the equal angles proportionals, they are similar to one another
(def. 1. 6,).

Next, Let it be required to describe upon a given straight line AB,
a rectilineal figure similar, and similarly situated to the rectilineal
figure CDKEF.

Join DE, and upon the given straight line AB describe the rectili*
neal figure ABHG similar, and similarly situated to the quadrilateral
figure CDEF, by the former case ; and at the points B, H in the
straight line BH, make the angle HBL equal to the angle EDK, and
the angle BHL equal to the angle DEK ; therefore the remaining
angle at K is equal to the remaining angle at L ; and because the
figures ABHG, (JDEF are similar, the angle GHB is equal to the
anerle FED, and BHL is equal to DEK ; wherefore the whole Angle
GHL is equal to the whole angle FEK ; for the same reason the an-
gle ABL is equal to the angle CDK : therefore the five-sided figuk^es
AGtlLB, CFEKD are equiangular; and because the figures AGHB,
CFED Are similar, Glf isto HB as Ft: to El) ; and as HB to HL,
so is ED to EK (4. 6.) ; therefore, ex ffiquali (22. 5.), GH is to HL,
as FE to P!K : for the same, reason, A is to BL, as CD to DK : and
BL isto LH, as (4. 6.) DK toKE, becau^se the riangles BLH, DKE
are equijinouUir : therefore, because the five-sided figures AGHLB,
OFEKD are equiangdlar, and have their sides about the equal aa«

149

SLEMENTB

gles proportionals, they are similar to one another ; and in the same
mariner a rectilineal figure of six, or more, sides may be described
upon a given straight line similar to one given, and so on. Which
was to be done.

PROP. XIX. THEOR.

SimUar frianghs are to one another in the duplicate ratio rfthe homar

logous sides.

Let ABC, DEF be simi-
lar triangles, having the an-
gle B equal to the angle E,
and let AB be to BC, as
DE to £F, so that the side
BC is homologous to EF
(def. 13. '5.) : the triangle
ABC has to the triangle
DEF, the duplicate ratio
of that which BC has to
EF.

Take BG a third proportional to BC and EF (II. 6.), or such that
BC : EF : : EF : BG, and join GA. Then, because
AB : BC : : DE . EF, alternately (16. 5,),
AB : DE : : BC : EF ; but
BC : EF : : EF : BG ; therefore (11. 5.)
AB : DE : : EF : BG ; wherefore the sides of the
friangles ABG, DEF, which are about the equal angles, are recipro--
cally proportional [ but triangles, which have the sides about two equal
angles reciprocally propor-
tional, are equal to one an-
other (15. 6.) : therefore
the triangle ABG is equal
to the triangle DEF ; and
because that BC is to EF,
as EF to BG ; and that if
three straight lines be pro-
portionals, the first has to
the third the duplicate ratio
of that which it has to the second ; BC therefore has to BG the du-
plicate ratio of that which BC has to EF. But as BC to BG, so is
(1. 6.) the triangle ABC to the triangle ABG : therefore the triangle
ABC has to the triangle ABG the duplicate ratio of that which BC
has to EF : and the triangle ABG is equal to the triangle DEF ;;
wherefore also the triangle ABC has to the triangle DEF the dupli-
cate ratio of that which BC has to EF. Therefore, similar triaa*
gles, 6lc. Q. E. D.
Cos. From this, it is manifest, that if three straight lines be pro^^

OF GEOMETRY. BOOK VI. 141

portionals, as the first is to the^ third, so is any triangle upon the first
to a similar, and similarly described triangle upon the second*

PROP. XX. THEOR.

/•

Similar polygons may he divided into the same number of similar tri»
angles, hauling the same ratio to one another that the polygons have ;
and the polygon^ have to one another the duplicate ratio of that which
their homologous sides have*

Let ABODE, F6HKL, be similar polygons, and let AB be the ho-
mologous side to FG : the polygons ABODE, FGHKL, may be di-
vided into the same number of similar triangles, whereof each has to
to each the same ratio which the polygons hare ; and the polygon
ABODE has to the polygon FGHKL a ratio duplicate of that which
the side AB has to the side FG.

Join BE, £^0, GL, LH : and because the polygon ABODE issimi-
lar to the polygon FGHKL, the angle BAE is equal to the angle GFL
(def. 1. a.), and BA : AE : : GF : FL (def. 1. 6.) : wherefore, be-
cause the tris^ngles ABE, FGL have an angle in one equal to an an-
gle in the other, and their sides about these equal angles proportion-
als, the triangle ABE is equiangular ,(G. 6.), and therefore similar, to
the triangle FGL (4. 5.) : wherefore the angle ABE is equal to the
angle FGL : and, because the polygons are similar, the whole angle
ABC is equal (def. 1. 6.) to the whole angle FGH ; therefore the re-
maining angle EEC is equal to the remaining angle LGH : now be-
cause the triangles ABE, FGL are similar,

EB : BA : : LG : GF ; and also because
the polygons are similar, AB : BC : : FG : GH (def. 1. 6.) ; there-
fore, ex aequali (22. 6.) EB : BC : : LG : GH, that is, the sides
about the equal angles EBC, LGH are_ proportionals ; therefore

f^^

^ H

(6. 6.) the triangle EBO is equiangular to the triangle LGH, and si.
mrlar to it (4. 6.). For the same reason, the triangle ECD is like-
wise similar to tho triangle LHK ; therefore the similar polygons
ABODE, FGHKL are divided into the same nuinber of similar tri.
angles.

142 " ELEMENTS

Also these triangles have, each to each, the same ratio which the
polygons have to one another, the antecedents being ABE, EBC,
£CD, and the consequents FGL, LGH, LHK : and the polygon
ABODE has to the polygon FGHKL the duplicate ratio of that
which the side AB has to the homologous side FG.

Because the triangle ABE is similar to the triangle FGL, ABE has
to FGL the duplicate ratio (19. 6.) of that which the side BG has to
the side GL : for the same reason, the triangle BEC has to GLH the
duplicate ratio of that which BE has to GL : therefore, ae the trian-
gle ABE to the triangle FGL, so (H. 5.) is the triangle BEG to the
triangle GLH. Again, because the triangle EBC is similar to the
triangle LGH, EBC has .to LGH the duplicate ratio of that which the
side EC has to the side LH : for the same reason, the triangle ECD
has to the triangle LHK, the duphcate ratio of that which EC has to
LH : therefore, as the triangle EBCtothe triangieLGH,so is (11. 5.)
the triangle. ECD to the triangle LHK : but it has been proved, that
the triangleEBC is likewise to the triangle LGH, asthe triangle ABE
to the triangle FGL. Therefore, as the triangle ABE is to the tri-
angle FGL, so is the triangle EBC, to the triangle LGH, and the tri-
angle ECD to the triangle LHK : and therefore, as one of the an-
tecedents to one of the consequents, so are all the antecedents to ali
the consequents (12. 5.). Wherefore, as the triangleABE to the tri-

angle FGL, so is the polygon ABCDE to, the polygon FGHKL : *ut
the triangle ABE has to the triangle FGL. the duplicate ratio of that
which the side AB has to the homologous side FG. Therefore also
the polygon ABCDE has to the polygon FGHKL the duplicate ratio
of that which AB has to the homologous side FG. Wherefore simi-
lar polygons, <&c. Q. E. D.

Cor. 1. In like manner it may be proved, that similar figures of
four sides, or of any number of sides, are one to another in the dupli-
cate ratio of their homologous sides, and the same has already been
proved of triangles : therefore, universally similar rectilineal figures
are to one another in the duplicate ratio of their homologous sides.

Cor. 2. And if to AB, FG, two of the homologous sides, a third
proportional M be taken, AB has (def. 11. 6.) to M the duplicate ra-
tio of that which AB has to FG : but the four-sided figure, or poly^
gon, upon AB has to the four-sided figure, or polygon, upon FG like-

OF GEOMETRY. BOOK VI. 149

Wise the duplicate ratio of that which AB has to FG : therefore, as
A B is to M, so is the figure upon AB to the fi^ufe upon FG, which
was also proved in triangles (Cor. 19. 6.). Therefore, universally,
it is manifest, that if three straight lines be proportionals, as the first
is to the third, so is any rectilineal figure upon the first, to a similar,
and similarly described rectilineal figure upon the second.

Cor. 3. Because all squares are similar figures, the ratio of any two
squares to one another is the same with the duplicate ratio of their
sides ; and hence, also, any two sin^ilar rectilineal figures are to oner
another as the squares of their homologous sides.

PROP. XXI. THEOR.

Rectilineal figures which are similar to the same rectilineal figure^ fwe

also similar to one another.

Let each of the rectilineal figures A, B be similar to the rectilineal
figure C : The figure A is similar to the figure B.

Because A is similar to C, they are equiangular, and also have their
sides about the equal angles proportionals (def. 1.6.). Again, because
B is similar to C, they are equiangular, and have their sides about the
equal angles proportionals (def. 1. 6.) : therefore the figures A, B,

are each of them equiangular to C, and have the sides about the
equal angles of each of them, and of C, proportionals. Wherefore
the rectilineal figures A and B are equiangular (1. Ax. 1.), and have
their sides about the equal angles proportionals (11. 5.). Therefore
A is similar (def 1. 6.) to B. Q. E. D.

PROP. XXII. THEOR.

If four straight lines be proportionals , the similar rectilineal figures
similarly described upon them shall also be proportionals ; and if the
similar rectilineal figures similarly described upon four straight lines
be proportionals, those straight lines shall be proportionals.

Let the four straight lines, AB, CD, EF, GH be proportionals, viz.
AB to CD, as EF to GH, and upon AB, CD let the similar rectilineal
figures KAB, LCD be similarly described ; and upon EF, GH the

144

ELEMENTS

similar rectilineal figures MF, NH, in like manner : the rectllineat
figure KAB is to LCD, as MFto NH.

To AB, CD take a third proportional (11. 6.) X ; and to EF, GH,
a third proportional O ; and because

AB : CD : : EF : GH, and
CD : X : : GH : (11. 5.)0,ex8Bquali (22. 5.)
AB : X : : EF : O. But
AB : X (2. Cor. 20. 6.) : : KAB : LCD ; and
EF : O : : (2. Cor. 20. 6.) MF : NH ; therefore
KAB : LCD (2. Cor. 20. 6.) : : MF : NH.

And if the figure KAB be to the figure LCD, as the figure MF to
the figure NH, AB is to CD, as EF to GH.

Make (12. 6.) as AB to CD, soEF to PR, and upon PR describe
(18. 6.) the rectilineal figure SR similar, and similarly situated to ei-

ther of the figures MF, NH : then, because that as AB'to CD, so is
EF to PR, and upon AB, CD are described the similar and similarly
situated rectilineals KAB, LCD, and upon EF, PR, in like manner,
the similar rectilineals MF, SR ; KAB is to LCD, as MF to SR ; but
by the hypothesis, KAB is to LCD, as MF to NH ; and therefore the
rectilineal MF having the same ratio to each of the two NH, SR, these
two are equal (9. 5.) to one another ; they are also similar, and simi^
larly situated ; therefore GH is equal to PR : and because as AB to
CD, so is EF to PR, and because PR is equal to GH, A B is to CD,
as EF to GH. If therefore four straight lines, <&c. Q. E. D. "^

PROP. XXIII. THEOR.

Equiangular parallelograms have to one another the ratio which is coni»

pounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms having the angle BCD
equal to the angle ECG ; the ratio of the parallelogram AC.to the
parallelogram CF, is the same with the ratio which is compounded
of the ratios of their sides.

Of GEOM^TkY. BdOk \l

Let 6C, CO be placed ia a straight lioe ; thi$tefi?)re DC and.CE ai«
«l3o in a straight line (1 4. L ) ^ complete the parallelogram DG ; and,
Uking any straiglit line K, make (12. 6.) as BC to CG^ K to L';
and as DC to CE, somake (12. 6.) L to M : therefore the ratios of K
to L, and L te M, are the same with the ratios of the sides, viz. of
9C to CG, and of DC to CE. But the ratio of K to M, is that whipk
IS said to be compounded (def. 10. 5.) of the ratios of K to L«^dt<
tovM ; wherefore also K has to M the a T\ TX

ratio compounded of the ratios of the - — — ^ ^

sides of the parallelograms. Now^
because as BC to CG, so is the paral-
lelogram AC to the pardlelogram CH
(1. 6.) ; 5ind as BC to CG, so is K
to L ; therefore K is (11. 5,) to L, ,
«s the parallelogram AC to the pa-
xallelcuram CH : again, because as
DC to CE, so is the parallelogram CH
to the parallelogram CF : and as DC
to CE, so is L to M ; therefore h is
(11. 5.) to M, as the parallelogram ^ j^ lyj-
CH to the parallelogram CF : there-
fore, since it has been proved, that as K to L, so is the parallelograni
AC to the parallelogram CH ; and as L to M, so the pahiUelogram
CH to the patallelogram CF ; ex «quali (22. 5.), £ ja^to M, ^a the
parallelogram AC to the parallelogram CF ; but K iias to ]H[ the ratio
"whuch is GQiBpoaQded of the ratios of the sid^s ; therefofe ^Uo the
parallelogram AC has to the parallelogram CF the ratio. which is
tsompounded of the ratios of the sides. Whecefiire equia^igulat
parailelogtaAis, &c. Q. £• D.

PROP, XXIV. THEORv

The paralktograms ahout the diameter of any paraUdogram, are tind-

lar to the whole, and to one another*

Let ABCD be a parallelogram, of which the dianleter is AC ; and
&G, HK the parallelograms about the diameter : the parallelograms
EG, HK are similar, both to the whole parallelogram ABCD, fmd to
lone another.

Because DC, GF are parallels^ the angle ADC is equal (20. l.)to
the angle AGF : for the same reason', because BC, £F are parallels^
the angle ABO is equal to the angle AEF : and each of the angles
BCD, EFG is equal to the opposite angle DAB (34. 11) alid therefore
are equal to one another, wherefore the parallelograms ABCD, AEFG
ve equiangular. And because the angle ABu is equal-to the aogld

146

ELEMENTS

AEF, and the angle BAG common to the
two triangles BAG, EAF, they are equian-
gular to one another ; therefore (4. 6.) as
AB to BG, so is AE to EF^; and because
the opposite sides of parallelograms are
equal to one another (34. !.)» AB is (7. 5.)
to AD, as AE to AG ; and DC to GB, as

GF to FE ; and also CD to DA, as FG to

GA : therefore the sides of the parallelo- j) "g^
grams ABCD, AEFG about the equal an-
gles are proportionals ; and they are therefore similar to one anothet
(def. 1. 6.) ; for the same reason, the parallelogram ABCD is similar
to the parallelogram FHCK. Wherefore each of the parallelograrns,
GE, KH is similar to DB : but rectilineal figures which are similar
to the same rectilineal figure, are also similar to one another (2K 6.) ;
therefore the parallelogram GE is similar to KH. Wherefore the
parallelograms, dec. Q. £. D.

PROP. XXV. PROB.

To describe a rectilineal figure which shall be similar to one, and eqwd

to another given rectilineal figure.

Let ABC be the given rectilineal figure, to which the figure to be
described is required to be similar, and D that^to which it must be
equal. It is required to describe a rectilineal figure similar to ABC,

and equal to D.

Upon the straight lineBC describe (cor. 45.1.) the parallelogram BE.
equal to the figure ABC ; also upon GE describe (cor. 45. 1 .) the paral-
lelogram CM equal to D, and having the angle FGE equal to the angle
GBL : therefore EC and GF are in a straight line (29. 1. 14. 1.), as also
LE and EM ; between BC and CF find (13. 6.) a mean proportional
GH, and upon GH describe (18. 6.) the rectilineal figure KGH simi-

lar, and similarly situated, to the figure ABC. And because BG is k>
GHas Gil to CF, and if three straight lines be proportionals, as the first
is to the third, so is (2. Gor. 20. 6.) the figure upoo the first to the «i-^

OF GEOMETRY. BOOK VI. 147

Milar and similarly decfcribed figure upon the second ; therefore as
BC to OF, so is the figure ABC to the figure KGH : but as BC to CF,
so is (1. 6.) the parallelogram BE to the parallelogram EF : therefore
as the figure ABC is to the figure KGH, so is the parallelogram BE to .
the parallelogram EF (11. 5.) : but the rectilineal figure ABC is equal
to the parallelogram BE ; therefore the rectilineal figure KGH is equal
(14. 5.) to the parallelogram EF : but EF is equal to the figure D ;
wherefore also KGH is equal to D ; and it is similar to ABC. « There-
fore the rectilineal figure KGH has been described similar to the
figure ABC, and equal to D. . Which was to be done.

PROP. XXVI. THEOR.

If two similar paraUdograme have a common angle, and he nmilarly

sUvatedy they are about the same diameter.

#
Let the parallelograms ABCD, AEFG be similar and similarly

situated, and have the angle DAB common ; ABCD and AEFG are
about the same diameter.
For, if not, let, if possible, the parallel- \ r^ Ti

ogram BD have its diameter AHC in adif. A- vr JL)

ferent straight line from aP, the diameter i
of the parallelogram EG, and let GF meet K^i
AHC in H ; and through H draw HK pa- E/
rallel to AD or BC ; therefore the paral- '
lelograms ABCD, AKHG being about the
same diameter, are similar to one another
(24. 6.) : ^herelfore, as DA to AB, so is !B
(def. 1. 6.) GA to AK; but because ABCD
and AEFG are similar parallelograms, as DA is to AB, so is GA to
AE ; therefore (11. 5.) as GA to AE, so GA to AK ; wherefore GA
has the same ratio to each of the straight lines AE, AK ; and conse-
quently AK is equal (9. 5.) to AE, the less to the greater, which is
impossible ; therefore ABCD and AKHG* are not about the same dia*
meter ; wherefore ABCD and AEFG must be about the same dia-
meter. Therefore, if two similar, dz;c. Q* E. D.

PROP. XXVn. THEOR.

Of all the rectangles contained by the segments of a given straight line^
tlie greatest is the square which is described on half the line.

Xict AB be a given straight line, which is bisected in C ; and let D
be any point in it, the squl^^ on AC is j- "q jp i — jg

greatcr-than the rectangl^ AD, DB.

For, since the straight Une AB is divided into two equal parts in C,
and into two unequal parts in D, the rectangle contained by AD and
DB, together with the square of CD, is equal to the square of AG

T49

ELEMENTS

(5, 2.). The square ef AC is therefore greater than the reettingT^
AD;I>B. Therefore, &c. Q. £. D.

PROP. XXVIII. PROB.

To ditide a gicen straight line, so thai the rectangle eoniained hy its
segments may he equal to a given space ; hut that space must not he
gweater than the square cf^ half the given line.

Let AB be the given -straight line^ and let the square upon the
given straight line C be the space to which the rectangle eontained
by the segments of AB must be equal, and this square, by the deter-
mination, is not greater than that upon half the straight line AB.

Biseet ABin D, and if the square upon AD be equal to the square
upon C, the thing required is done : But if it be not equal to it, AH
must be greater than C, according
to the determination : Draw DE at
right asglea to AB, and make it
equal to C : produce ED to F, so
that EF be equal to AD or DB, and
from the centre £, at the distance
EF,. describe a circle meeting AB
in 6. Join EG ; and because AB.
is divided equally in D, and un-
equally in G, AG.GB + DG« = (5w 2.) DB« = EG'. But (47. I.)
ED'+DG»=EG'; therefore AG.GB+DG»=Eiy+DG\and tak.
ing away D^*, AG.GB— ED^. Now ED=C, therefore the rectan.
gle AG.GB is equal to the square of C : and the given line AB i»
divided in G, so that the rectangle contained by the segments AG>
GB is equal to the square upon the given straight line C. Which wa^^^
to be done.

PROP. XXrX. PROB.

To produce a gicen straight line, so that the rectangle contained by tA»
segments hetween the extremities of the given line, and the pohit tOr
which it is produced, may he equal to a given space*

Let AB be the given straight line, and let the square upon the
given straight line O be the space to which the rectangle under the
segments of AB produced, must be equal.

Bisect. AB in D, and draw BE at right angles to it, so that BE be
equal to C ; aiid having joined DE, from the centre D at the distance

OF GEOMETRY. BOOK VL

DE describe a circle meeting AS
produced in G. And because AB is
bisected in D, and produced to G,
(6. 2.) AG.GB+DB»=DG«=DE».
But (47. 1.) DE» =DB^ + B£>,
therefore AG.GB + DB» = DB« +
BE» and AG.GB=BE». Now, BE
=xC^ wherefore the straight line AB
is produced to G, so that the rectan-
gle contained by the segments AG,
GB of the line produced, is equal to the square of C. Which waa
to be done.

PROP. XXX. PROB.

To cut a given straight line in extreme and mean raiUK

Let AB be the given straight line ; it is required to cut it in extreme
aiid mean ratio.

Upon AB describe (46. 1.) the square BC, and produce CA to D,
so that the rectangle CD. DA may be equal to the square CB (29. 6.)*
Take AE equal to AD, and complete the rectangle DF under DC and
AE, or under DC and DA. Then, because ...
the rectangle CD. DA is equal to the square ^
CB, the rectangle DF is equal to CB. Take
away the common part C£ from each, and
the remain4er FB is equal to the remainder a
DE. But FB is the rectangle contained by
FE and £B, that is, by AB and BE ; and DE
is the square upon AE ; therefore AE is a
mean proportional between AB and BE (17.
6.), or AB is to AE as AE to EB. But AB
is greater than AE ; wherefore AE is great- ^
er than EB (14. 5u) : Therefore the straight C
line AB is cut in. extreme and mean ratio in
£ (def. 3. 6.}. Which was to be done.

Otherwise.

Let^AB be the given straight line ; it is required to cut it in ex-
treme and mean ratio.

Divide AB in the point C, so that the rectangle contained by AB,

BC be equal to the square of AC (11.2.): •

Then because the rectangle AB.BC is equal A - C B

to the square of AC, as BA to AC, so is AC

ISO

ELEHENTS

to CB (17. 6.) ; Therefore AB is cut in extreme aiMl mea,n ratio in C
(def. 3. 6.). Which was to be done.

PROP. XXXI. THEOR.

In right angled Irianglesj the rectilineal figure described upon the side
opposite to the rigfu angle, is equal to the similar , and similarly de^
cribed figures upon the sides containing tJie right angle.

Let ABC be a right angled triangle, having the right angle BAG :
The rectilineal figure described upon BC is equal to the similar, and
similarly described figures upon BA, AC.

Draw the perpendicular AD ; therefore, because in the rignt an-
gled triangle-ABG, *AD is drawn from the right angle at A perpendi.
cular to the base BC, the triangles ABD, ADC are similar to the
whole triangle ABC, and to one another (S. 6.), and because the tri-
angle ABC is similar to ADB, as CB to BAjSo is BA toBD (4. 6.) ;
and because these three straight lines are proportionald, as the first to
the third, so is the figure upon the first to the similar, and similarly
described figure upon the second (2. Cor. 20. 6.) : Therefore, as CB
to BD, so is the figure upon CB to
the similar and similarly described
figure upon BA : and inversely
(B. 5.), as DB to BC, so is the
figure upon BA to that upon BC ;
for the same reason as DC to CB,
€0 is the figure upon CA to that ^-
upon CB. Wherefore, as BD and ^
DC together to BC, so are the
figures upon BA and on AC, toge-
ther, to the figure upon BC (24. 5.) ; therefore the figures on BA, and
on AC, are^ together equal to that on BC ; and they are similar
figures. Wherefore, in right angled triangles, &c. Q. E. D.

PROP. XXXII. THEOR.

If t%D0 triangles, which have two sides of the one proportional to two
sides of the other, he joined at one angle, so as to have their homolo^
gous sides parallel to one another ; their remaining sides shall he in
a straight line* '

' Let ABC, DCE be two triangles which have two sides BA, AC
proportional to the two CD, DE, viz. BA to AC, as CD to DB ; and
let AB be parallel to DC, and AC to D£ ; BC and CE are in a
straight line.

OP GEOMETRY. ROOK VL 151

Because AB is parallel to DC, and the straight line AC meets them,
the alternate angles BaC, A CD are equal (29. 1.) ; for the same
reason, the angle CDE is equal to the angle ACD ; wherefore also
BAC is equal to CDE : And because the triangles ABC, DC£ have
one angle at A equal to one at D, ^^
and the sides about these angles
proportionals, vi2, BA to AC, as
CD to DE, the triangle ABC is
equiangular (6. 6.) to DCE :
Therefore the angle ABC is
equal to the angle DCE : And the
angle BAC was proved to be
equal to ACD : Therefore the
whole angle ACE is equal to the two angles ABC, BAC ; add the
common angle ACB, then the angles ACE, ACBare equal to the- an-
gles ABC, BAC, ACB : But ABC, BAC, ACB are equal to two right
angles (32. 1.) ; therefore also the angles ACE, ACB are equal to
two right angles : And since at the point C, in the straight line AC,
the two straight lines BC, CE, which are on the opposite sides of it,
make the adjacent angles ACE, ACB equal to two right angles ;
therefore (i4. 1.) BC and CE are in a straight line. Wherefore, if
two triangles, d^c. Q. E. D. •

PROP. ^XXIII. THEOR.

In eqtud circles, angles, whether at the centres or circumferences, have
the same ratio which the arches, on which they stand, have to one an'
other : So also have the sectors.

Let ABC, DEF be equal circles ; and at their centres the angles
BGC, EHF, and the angles BAC, EDF at their circumferences ; as
the arch BC to the arch EF, so is the angle BGC to the angle EHF,
and the angle BAC to th\e angle EDF : and also the sector BGC to
the sector EHF.

Take any number of arches CK, KL, each equal to BC, and any
number whatever FM, MN each equal to EF; and join GK, GL, HM,
HN. Because the arches BC, CK, KL are all equal, the angles BGC,
CGK, KGL are also all equal (27. 3.) : Therefore, what multiple
soever the arch BL is of the arch BC, the same multiple is the angle
BGL of the angle BGC : For the same reason, whatever multiple the
arch EN is of £e arch EF, the same mp^^.tiple is the angle EHN of the
angle EHF. But if the arch BL, be equal to the arch EN, the angle
BGL is also equal (27. 3.) to the angle EHN ,- or if the arch BL be
greater than EN, likewise the angle BGL is greater than EHN : and
if less, less : There being then four magnitudes, the two arches, BC,
£F, and the two angles BGC^ EHF, ernd of the arch BC, and of the

lit

^LiiM^iixmL

angle BCtC, hare been taken any equimultiples whatever, viz* the
arch BL, and the angle BGL ; and of the arch EF, and of the angle
EHF, any equimultiples whatever, viz. the arch EN, and the angle
EHN : And it has been proved, that if the arch BL be greater than
EN, the angle BGL is greater than EHN ; and if equal, equal ; and
if less, less ; As therefore, the arch BC to the arch EF, so (def. 5.
5.) is the angle BGC to the angle EHF : But as the angle BGC is
to the angle EHF, so is (15. 5.) the angle BAG to the angle EDF,
for each is double of each (20. 3.) : Therefore, as the circumference
BC is to EF, so is the angle BGrC to the angle EHF, and the angle
BAC to the angle EDF.

Also, as, the arch BC to EF, so is the sector BGC to the sectol*
EHF. Join BC, CK, and in the arches BC, CK take any points X,
O, and join BX, XC, CO, OK : Then, because in the triangles GBC^
GCK, the two sides BG, GC are equal to the two CQ, GK, and also
contain equal angles ; the base BC is^equal (4. 1.) to the base CK,
and'the triangle GBC to the triangle GCK : And because the arch
BC is equal to the arch CK, the remaining part of the whole circum-
ference of the circle ABC is equal to the remaining part of the whole
circumference of the same circle : Wherefore the angle BXC is
equal to the angle COK (27. 3.) ; and the segment BXC is therefore
similar to the segment COK (def. 9. 3.) ; and they are upon equal
straight lines BC, CK : But similar segments of circles upon equal
straight lines are equal (24. 3.) to one another : Therefore the seg-
ment BXC is equal to the segment COK : And the triangle BGC is
equal to the triangle CGK ; therefore the whole, the sector BGC is
equal to the whole, the sector CGK : For the same reason, the sec-i
tor KGL is equal to each of the sectors BGC, CGK; and in the same
manner, the sectors EHF, FHM, MHN, may be proved eqUal to one
another : Therefore, what multiple soever the arch BL is of the arch
BC, the same multiple is the sector BGL of the sector BGC. Fox'
the same reason, whatever multiple the arch EN is of EF, the
same multiple is the sector EtiN of the sector EHF ; Now if the

OP GEOMETRT. BOOK VI. 163

irch BL be equal to EN, the sector B6L is equal to the sector EHN ;

C F

«ind ifthe arch BL be greater than EN, the sector BGL is greater thtii
the sector EHN ; and if less, less : Since, then, there are four mag*
nitudes, the two arches BC, £F, and the two sectors BGO, £HF,and
t>f the arch BC, and sector BGC, the arch BL and the sector BGL
are any equimultiples whatever ; and of the arch EF, ahd sector
£HP, the arch EN and sector EHN, are any equimultiples whatever i
and it has been proved, that if the arch BL be greater than £N, the
•sector BGL is greater than the sector EHN ; if equal, equal ; and
if less, less ; therefore {ded 5. 5.) as the arch BC, is to the arch EF,
so is the sector BGC to the sector EHF. Wherefore, in equal cir-
cles, d2;c. Q« E. D.

PROP. B. THEOR.

JjT an angle of a triangle be bisected by a slraight Kite, itikick likewtst
cuts the base ; the rectangle contained by the sides of the triangle is
equal to the rectangle contained by the segments of the base^ together
iffUh the square of the straight line bisecting the angle*

Let ABC be a triangle, and let the angle BaC be bisected by the
Bt^ight. line/ AD ; the rectangle BA.AC is equal to the rectangle
Bd.DC, together with the square of AD.

Describe the circle (5. 4.) ACB about
the triangle, and produce AD to the cir« ^^,,^». .<A.

cumference in E, and join EC. Then,
because the angle BAD is equal to the an*

ffle CAE, and the angle ABD to the angle j^/->^ / ^Q

\2L 3.) A£C, for they are in the same
segment ; the triangles ABD, AEC are
equiangular to one another : Therefore
BA : AD : : £A : (4. 6.) AC, and conse-
quently,* BA.AC = (16. 6.) AD.AE ==
ED.DA (3. 2.) +DA>. But ED.DA=
BD.DC, therefore BA.AC=BD.DC+
DA^. Wherefore, if an angle, &c. Q. £. D.

164

ELEMENTS
PROP. C. THEOR.

If from any angle of a triangle a straight line he drawn perpendicular
to the base ; the rectangle contained by the sides of the triangle is
equal to the rectangle contained by the perpendicular ^ and the diame^
ter of the circle described about the triangle*

Let ABC be a triangle, and AD the perpendicular from the angle
A to the base BC ; the rectangle BA«AC is equal to the rectangle
contained by AD and the diameter of the circle described about the
triangle.

Describe (5. 4.) the circle ACB about
the triangle, and draw its diameter AE,*
and join £C ; Because the right angle
BDA is equal (3. 3.) to the angle EC A
in a semicircle, and the angle ABDto jg
the angle AEC, in the same segment
(21. 3.) ; the triangles ABD, AEC are
equiangular : Therefore, as (4. 6.) BA
to AD, so is EA to AC : and conse-
quently the rectanglip BA.AC is equal
(16. 6.) to the rectangle EA.AD. If,
therefore^ from an angle, dec. Q. E. D.

PROP. D. THEOR.

The rectangle contained by the diagonals oj a quadrilateral inscribed
in a circle, is equal to both the rectangles, contained by its opposite
sides.

Let ABCD be any quadrilateral inscribed in a circle, and let AC.
BD be drawn ; the rectangle AC.BD is equal to the two rectangles
AB.CD, and AD.BC.

Make the angle ABE equal to the angle DBC ; add to each of
these the common angle EBD, then the angle ABD is equal to the
angle EBO : And the angle BDA is equal to (2L 3.) the angle BC£,
because they are in the same segment ;
therefore the triangle A BD is equian-
gular to the triangle BCE. Wherefore
(4. 6.), BC : CE : : BD : DA, and con-
sequently (16. 6.) BC.DA=--BD.CE.
Again, because the angle ABE is equal
to the angle DBC, and the angle (21. '3.)
BAE to the angle BDC, the triangle ABE
is equiangular to the triangle BCD ;
therefore BA : AE : : BD : DC, and
BA.DC^BD.AE : But it was shewn
that BC.DA=:BD.CE ; wherefore BC.

OF GEOMETRY. BOOK VI. 165

DA +BA.DC=BD CE+BD.AE=BD.AC(1.2.). Thati8,thercct.
angle contained by BD and AC, is equal to the rectangles contained
by AB, CD, and AD, BC. Therei'ore the rectangle, ~&c. Q. E. D.

PROP. E. THEOR.

If an arch of a circle he bisected^ and from the extremities of the archf
and from the point of bisection, straight liiies he drovn to any point
in tlie circumference, the sum of the iwo linrs drown from the txtre^
mities of the arch witl have to the line ^ravm from the jtoint of hu
section, the same ratio which the straight line subtending tJie arch
has to tlie- straight line subtending half the arch.

Let ABD be a circle, of which A B is an arch bisected in C, and
from A, C, and B to D, any point whatever in the circumference, let
AD, CD, BD be drawn ; the sum of tlie two lines AD and DB ha»
to DC the same ratio that BA has to
AC.

For since ACBD is a quadrilateral
inscribed in a circle, of which the di-
agonals are AB and CD, AD.CB+
DBiAC(D. 6.)=AB.CD; but AD.
CB + DB.AC=:AD.AC + DB.AC,
because CB=AC. Therefore AD*
AC+DB.AC, that is (1. 2.), (AD+
DB) AC= AB.CD. And because the
sides of equal rectangles are recipro-
cally proportional (14. 6.), AD+DB :
DC : : AB : AC, Wherefore, &c.
O- E. D.

PROP. F. THEOR.

Jfiwo points he taken in the diameter of a circle^ such that the rectan^
gle contained hy the segments intercepted, between them and the cen»
ire of the circle be equal to the square of the radius : and if from
the^e. 'points two straight lines be drawn to any point whatsoever in
the circumference of the circle, the ratio of these lines will be the
same with the ratio of the segments intercepted between the two first
mentioned points and the circumference of the circle.

Let .-VBC be a circle, of which the centre is D, and in DA produc-
ed, let the points E and F be such ihHt the rectangle ED, DF is equal
to th'e square of AD ; from E and F to anv point B in the circumfer^
ence, let EB, FB be drawn ; FB : BE : TFA : AK.

Join BD, and because the rectangle FD, DE is equal to the square
of AD^ that is, of DB, FD : DB : : DB : DE (17. G.).

The two triangles, FDB, BDE have therefore the sides proportion-
al that are about the common angle D; therefore they are equiangu.
lar (6. 6.), the angle DEB being equal to the angle DBF, and DBE to
DFB. Now since the sides about these equal angles are also propor«

ELEMENTS

fional (4. 6.), FF : BD : r BE i ED, and alternately (16. 5.), FEj
BE ; : BD : ED, oir FB : BE ; : AD : DE. But because FD : DA
: : DA : DE, by division (17. 5.), FA : DA ; : AE : ED, andaher^
nately (11. 5.) FA : AE : : DA : ED. Now it has been shewn 4hat
FB : BE : : AD : DE, therefore FB : BE : : FA : AE. Therefore,,
4ic. Q. E. D.

CoK. If AB be drawn, because FB : BE : : FA : AE, the angle
FBE is bisected (3. 6.) by AB. Also, since FD : DC : : DC : DE,
by conrtposition (18. 6.), FC : DC : : CE : ED, and since it has becB
•hewn that FA : AD (DC) r : AE : ED, therefore, ex aequo, FA r
AE : : FC : CE. But FB : BE : : FA : AE, therefore, FB : BE : i
FC : CE (11. 5.) so that if FB be produced to G, and if BC be
drawn, the angle EBG is bisected by the line BC (A. 6.).

PROP. G. THEOR.

If from the extremity of Ihe didmeler of a circle a straight line he drenon
in the circte^ and if either within the circle or produced without it, i$
meet a line perpendicular tathe same diameter , the rectangle con*
tained by the straight line drawn in the cirdCj and the segment of it,
intercepted between the extremity of the diameter and the perpendicuf
lary is equal to the rectangle contained by the diameter and the seg-
ment flfit cut off by the perpendicular.

Let ABC be a circle, of which AC is a diameter, let DE be pet*

OF GEOMETRY. BOOK VL

157

pendkilarto the diameter AC, and let AB meet DE m F; the Feci.
angfeBA.AF is equal to the rectangle CA.AD. Join BC, and be-
causeABC is an angle in a semicircle, it is a right angle (3^. 3.) :
Now,the angle ADF is'also a right angle (Hyp.) ; and the angle
BACs either the same with DAF, or vertical to it ; therefore the
trianges ABC, ADF are equiangular, and BA : AC : : AD : AF
(4. 6. ,' therefore also the rectangle BA.AF, contained by the ex-
treme:^ is equal to the rectangle ACAD contained by the means
(16. 6^. If therefore, &c. Q. E. D.

PROP. H. , THEOR.

The p^mdiculars drama from the three angles of any triangle to the
oppQsUe sides intersect one another in the same point.

Let ABC be a triangle, BD and CE twp perpendiculars intersect-
iog uod another in F ; Let AF be joined, and produced if necessary^
let it ueet BC in G, AG is perpendicular to BC.

Joii DE, and about the triangle. AEF let a circfe be described,
ASF then, because AEF is a right angle, the circle described aboul
the tritngle AEF will have AF for
its diameter (3L 3.). In the
same manner, the circle describ-
ed about the triangle ADF .has
AF for its diameter ; therefore
the points A, E, F and D, are in
the circumference of the ^ame
circle. But because the angle
EFB is equal to the angle DPC
(15. l.j, and slso the angle BEF
to the angle CDF, being both right
angles, the triangles BEF, and
CDF are equiangular, and there-
fore BF : EF : : CF : FD (4. 6.),
or Alternately (16. 6.) BF : FC

: : EF : FD. Since, then, the sides about the equal angles BFC,
EFD are proportionals, the triangles BFC, EFD are also equiangular
(6. 6.) ; wherefore the angle FCB is equal to the angle EDF. But
EDF is eqvial to EAF, because they are angles in the same segment
(21. 3.).; therefore the angle EAF, isequal to the angle FOG: Now,
the angles AFE, CFG are also equal, because they are vertical an-
gles; therefore the remaining angles AEF, FGC are also equal (32.
1.) : But AEF is a right angle, therefore FGC is a right angle, and
AG is perpendicular to BC. Q. E. D.

CoR. The triangle AD E is similar to the triangle ABC. For the
two triangles BAD, CAE having the angles at D and E right angles,
and fhe angle at A common, are equiangular, and therefore BA : AD
: : CA : AE, and alternately BA : CA : : AD : AE ; therefore the two
triangles BAG, DAE, have the angle at A common, and the sides about

158

ELEMENTS, &e.

that angle proportionals, therefore they are equiangular (6. I j and
similar.
Hence the rectangles BA.AE, CA. AD are equal.

PROP. K. THEOR.

If from any angle of a triangle a perpendicular he dravm to th oppo-
site or base : the rectangle contained by the sum and differnce of
the other two sides, is equal to the rectangle contained by $e sum
. and difference of the segments, into which the base is divider by the
' perpendicular.

Let ABC he a triangle, AD a perpendicular drawn from he an-
gle A on the base EC, so that BD, DC are the segments of the base ;
(AC+AB) (AC— AB; = (CD+DB) (CD— DB.)

From A as a centre with the radius AC, the greater of the two
sides, describe the circle CFG : produce AB to meet the circumfer-
ence in E and F, and CB to meet it in G. Then because AF=AC,
BF = AB-f AC, thesumof thesides; and since AEs=AC,BE = AC,
— AB= the difference of the sides. Also, because AD drawn frora
the centre cutsGC at right angles, it bisects it ; therefore, when tho
perpendicular falls within the triangle, BG = DG— DB = DC— DB=
the difference of the segments of the base, and BC=:BD+DC= the
sum of the segments. But when AD falls without the trian/sjle, BG =
DG-f DB=CD-|-DB= the sum of the segments of ihe base, and BC
—CD - DB= the difference of the segments of the baJ?e. Now, in
both cases, because B is the intersection of the tw^) lines FE, GC»
^T^"! \"oh .''i!''*^' FB.BE=CB.BG; that is^s has been shewn,
(AC+AB) (AC^ABJ==(CD+DB) (CD~DB). Therefore, &c.

< ^

■^' " »-

SUPPLEMENT

TO THE

ELEMENTS

GE OME TRY.

%•

ELEMENTS

GEOMETRY.

SUPPLEMENT.

■ ■■*■ ■■ ■■^■■i» I ^-

BOOK I.

OF THE QUADRATURE OF THE CIRCLE.

t I mm ^1 il.

DEFINITIONS.

A CfiORD of an a^h of a circle is the straight line Joining th& 6ktr«itl(«
ties of the arch ; or the straight line which subtends the a]fch4

The petimet^r of any figure is the length of the line or lines, by which
it is boundedk

lit.

The area of any figure is tiie space contained within it«

AXIOM*

The least line that tan be drawn betWeeti two poibtd, is a straight \iti6 1'
and if two figures have the same straight line for their base, that
which is contained within the other, if its bounding line or lines bd
not any wh^re conveic towards the base, has the leiEust peHdietei'.
CoRk 1. Hence the perimeter of any polygon inscribed in a circle
is less than the circumference of the circle.

'Cor. 2. If from a pd^ni two straight lines be drawn touching a cir'
de, these two lines are together greater than the arch intercepted be"
tween them ; and hence the perimeter of any polygon described
about a circle is greater than the circumference of the circle^

162

SUPPLEMENT TO THE ELEMENTS

PROP. L THEOR.

fffrom the greater of two unequal magnitudes there he taken away its
hdlfy and from the remdind^ its. half; and so on ; There wUl at
length remain a magnitude less than the least of the proposed mag*
nUudes*

Let AB^and C be two unequal magmtudes, of which AB is the

•F

greater: If from AB there be taken away its half,
and from the remainder its half, and so on ; there a
shall at length remain a magnitude less than C.

For C may be muTtipIiecl so as, at length, to be.
come greater than AB. Let DE, therefore, be a .«^
multiple of C, which is greater than AB, and let ^^
it contain the parts DF, FG; G£, each equal to
C. From AB take BH equal to its half, and from ^_ ^
the remainder AH, take HK equal to its half, and
so on, until there be as many divisions in AB as
there are in DE ; And let the divisions in AB be
AK, KH, HB. And because DS is greater than
AB, and EG taken from- DE is not greater than
its. half, but BH taken from AB is equal to its
half; therefore the remainder GD is greater than ]^
the remainder HA.r Again, because GD.is great-
er than HA, and GF is not greater than the half of GD, but HK is
equal to the half of HA ; therefore the remainder FD is greater than
the remainder AK. And FD is equal to C, therefore C is greater
than AK ; that is, AK is less than C. Q. E. D.

PROP. II. THEOR. -

Equilateral polygons, of the same number of sides, inscribed in drdes,
are similar ^ amd are to one another as the squares of the diameters of
the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the
same number of sides inscribed m the circles ABD and GHK ;
ABCDEF and GHIKLM are similar, and are to one another as the
squares of the diameters of the circles ABD, GHK.

Find N and O the centres of the circles, join AN aiid BN, as also
GO and HO, and produce AN and GO till they meet the circumfer-
ences in D and K. .

OP GEOMETRY. BOOK I.

168

Because the straight lines AB, BC, CD, DE, EF, FA, are all equal,
the arches AB, BC, CD, DE, £F, FA are also equal (28. 3.). For
the same reason, the arches GH, HI, IK, KIj, LM, MG are all equal,
and they are equal in number io the others; therefore; whatever
part the arch AB is of the whole circumference, ABD, the same is
the arch GH of the Qircumference GHK. But the angle ANB is the
same part of four right angles, that the arch AB is of the circumfer-
ence ABD (33. 6.) ; and the angle GOH is the same part of four
right angles that the arch GH is of the circumference GHK (33. Gi)
therefore the angles ANB, GOH are each of them the same part of
four right angles, and therefore they are equal to one another. The
isosceles trian^lea ANB, GOH are therefore equiangular (6. 6.}, and
the angle ABN equal to the angle GHO ; in the same manner, by
joining NC, 01, it may be proved that, the angles NBC, OHI airet

equat to one another, and to the angle ABN. Therefore the Whol^^
angle ABC is equal to the whole GHI ; and the same may be proved
of the angles BCD, HIK, and of the rest. Therefore, the polygons
ABCDEF and GHIKLM are equiangular to one another ; and since
they are equilaferal, the sides about the equal angles are proportion-
nls ; the polygon ABCD is therefore similar to the polygon GHIKLM
(d^f. L 6.). And bdeause similar polygons are as the squares ot

IM ^ SUPPLEMENT TO THE ELEMENTS

tb'eir homologous sides (20. 6.)) the polygon ABCDEF is to the p^
lygoD GHIKLM as the square of AB to the square of GH ; but be-
cause the tfiangles ANB, GOH are equiangii^r, the sqaare of AB 19
to the square cfGH as the square c^ AN to the square of GO (4. 6.)
or as four times the square of AN tafour times the square (15. 5.)
of GO^ that is, as the square of AD to. the square of GK (2. Cor. 8^.
S.). Therefore also^ the polygon ABGDEF rs to the polygon
GHIKLM as the square of AB to the square of GK ; and they have
also been' shewn to be similar. Therefore, &c. Q. E. D.

Cor. Every equnateral polygon inscribed in a cirele is also, eqiii*
angular : For the isosceles triangles, which have their commea ver-
tex in the eentre, are all equal and similar ; therefore, the angles at
their bases are all equal, and the angles of the polygoa Vite therefoiei
also equaL

PROP. Ifl. THEOH.

The sidle of on^ equHalercA polygon xTiscrihed in a circle being gwen^ to
JUid the McZe of a polygon of the same number of sides^ described c^bouf
the circle.

Let ABCDEF be an equilateraf polygon inscribed in the circlie
ABD ; it is required to find the side of an equilateral polygon of the
same number of sides described about the^ circle.

Find G the centre of the circle ; join GA, GB, bisect the arch AB
in H ; and through H draw KHL touching the circle in H, and meet-
ing GA and GB produced in K and L ; KL is the side of the polygon
required.

Produce GF to N, so that GrN may be equal to GL ; join KN, and
from G draw GM at right angles to KN, join also- HG..

Because the arch AB is bisected in H,, the angle AGH is eqjoal fei
the angle BGH (27. 3.) ; and
iecause KL touches the circle
Ri H„the angles LHG, KHGare
right an^es (16. S^) ; therefore,
there are two angles of the tri-
angle HGK, equal to two an-
gles of the triangle HGL, each
to each* But the side GH is
common to these triangles ;
therefore they are equal (26.
I.) J and GL is equal to GK.
Again, in the triangles KGL,
KGN, because GN is equal to
GL ; andGK common, and also
the angle LGK equal to the an*
gle KGN; therefore the base KL is equal to the base KN (4.-l.>.
j&ui because the triangle KGN is, isosceles, the angle GKN is equal

OF GEOMETRY. BOOK I. 165

to the angle GNK, and the angles GMK, GMN aire both right angles
by coDstructioQ ; wherefore, the triangles GMK, GMN have two an-
gles of the one equal to two angles of the other, and they have also
the side GM conunon, therefore they are C|qual (26. 1.), and the side
KM is equal to the side MN, so that KN is bisected in M. But KN
is equal to KL, and therefore* their halves KM aqd KM are also
equal. Wherefore^ in the triangles GKH, GKM, the two sides GK
and KH are equal to the two GK and KM, each to each ; and the
angles GKH, GKM, are also equal, therefore GM is equal to GH (4.
1.) \ wherefore, the point M is in the circumference of the circle ;
and because KMG is a right angle, KM touches the circle. And in
the same manner, by joining, the centre and the other angular points
of the inscribed polygon, an equilateral polygon may be described
about the circle, the sides of which will each be equal to KL, and
will be equal in number to the sides of the inscribed polygon. There-
fore, KL is the side of an equilateral polygon, described about the
circle, of the same number of sides with the inscribed p<^ygon
ABCDEF ; which was to be found.

Cor. 1. Because GL, GK, GN, and the other straight lines drawn
from the centre G to the angular points of the polygon described about
the circle ABD are 'all equal ; if a circle be described from the centre
G, with the distance GK, the polygon will be inscribed in that cir-
cle ; and therefore it is similar to the polygon ABCDEF (2. 1.).

Cob. 2. It is evident that AB, a side of the inscribed polygon, is to
KL, a side of the circumscribed, as the perpendicular from G upon
AB, to the perpendicular from G upon KL, that is, to the radius of the
circle ; therefore also, because magnitudes have the same ratio with
their equimultiples (15. 5.), the perimeter of the inscribed polygon is
to the perimeter of the circumscribed, as the perpendicular from the
centre, on a side of^the inscribed polygon, to the radius of the circle*

PROP. IV. THEOR.

A circle being giverij two similar ^polygons may he founds the one de^
scribed about the circle^ and the other inscribed in i/, which shall dif*
ferfrom one another by a space less than any given space.

Let ABC be the given circle, and the square of D any given space ;
a polygon may be inscribed in the circle ABC, and a similar polygon
described about it, so that the difference between them shall be less
th|Ln the square of D.

In the circle ABC apply the straight line AE equal to D,and let AB
be a fourth part of the circumference of- the circle. From the cir^
cumfereiice AB take awity its half, and from tba remainder its half, and
so on till the circumference AF is found less than the circumference
AE (1. 1. Sup.). Find the centre G ; draw the diameter AC, as also
the straight lines AF and FG ; and having bisected the circunifl^rence

Id$ . SUPPLEMENT TO THE ELEMENTS

AFin K, join K€r, and draw HL touching the circle in K, and meet^
ing GA and GrF produced in H and L ; join CF.

Because the isosceles triangles HGL and AGFhayelhe common
angle AGF, they are equiangular (6. 6.) and the ungles GHK, GAF
are therefore equal to one another. But the angle GKH, €FA are-
also equal, for they are right angles ; therefore the triangled HGK,.
AOF, are likewise equiangular (32. 1.).

And because the arch AF was found by taking from the arch AB ita
half, and fiom that remainder itahalf> and so on, AF will be contained
a certain number of times, exactly, in the arch AB, and therefore it
will alfto be contained a certain nUn]Gi>er of times, exactly, ia the whole^
circumference, ABC ; and the straight line AFis therefore the side of
an equilateral polygon inscribed in the circle ABC. Whereibre tilso,.
HL is the side of an equilateral polygon, of the same number of sides^
described about ABC (3. L Sup.). Let the polygon described about
t^ circle be called M, and the polygon inscribed be called N ; theto,^

because these polygons are similar (Cor. 3. 1.) they are as the square*
t)f the homologous sides HL and AF (Sup. 3. Cor. 20. 6.), that is^ be-
i5ause the triangles HLG, AFG are similar, as the square of HG ta
the square of AG, that is of GK. But the triangles HGK, ACF have
been proved to be similar, and therefore the square of AC is to the
square of CF as the polygon M to the polygon N ; and, byconversion,
the square of AC is to its excess above the squares of CF, that is, to the
square of AF (47. 1 .), as the polygon M to its excess above the poly-
gon N. But the square of AC, that is, the square described about:
the circle ABC is greater than the equilateral polygon of eight sides:
described about the circle, because it contains that polygon ; and, for
the same reason, the polygon of eight sides Is greater than the poly-
gon of sixteen, and so on ; therefore, the square of AC is greater
than any polygon described about the circle by the continual bisectioa
#f the ar^ AB ; it is Uierofore greater than the polygon M. Now, it

OP GEOMETRY. BOOK t.

hafl! been demoDstrated, that the square of AC i» to the squaie of AP
us the polygon M to the difference of tke polygons; thei«fpre, since
the square of AC is greater ths^u M , the square of AF is greater thait
the difference of the polygons (14. 5.). The difference of the poly-
gons is therefore less than the square of AF ; but AF is less than D $
therefore the difference of the polygons is less than the square of D ;
that is, than the given space. Therefore, i&c. Q. E. D.

Cor. 1. Because the polygons M and N differ from one another
more than either of them differs from the circle, the difference between
each of them and the circle is less than the given space> viz. the square
of D. And therefore, however small any given space may be, a poly-
gon may be inscribed in the circle, and another described about it, each
of which shall differ from the circle by a space less than the given space.

Cor. 2. The space B, which is greater than any polygon that oan
be inscribed in th^ circle A, and less than any polygon that can be
described about it, is equal to the circle A. If not, let them be un-
equal ; and first, let B exceed A by the spacq C. Then, because the
polygons described about the circle A are all greater than B, by hy.
pothesis ; and becanse B is greater than A by the space C, therefore
no polygon can be described about the circle A, but what must exceed
it by a space greater than C, which is absurd. ^ In the same manner,
if B be less than A by the space O, it is shewn that no polygon can
be inscribed in the circle A, but what is less than A by a space great-
er than C, which is also abdurd. Therefore, A and B are not un-
equal ; that is, they are equal to one another.

PROP. V. THEOR.

The area of any circle is equal to the rectangle contained fty the-Mcmi*
' diameter J and a straight line equal to half the circumference.

Let ABC be a circle of which the centre is D, ^nd the diameter
AC ; if in AC produced there be taken AH equal to half the circum-
ference, the area of the circle is equal to the rectangle contained by
DA and AH.

les

SUPPLEMENT TO THE ELEMENTS .

Let AB be the side of any equilateral polygon inscribed in the cir^
de ABC ; bisect the circumference AB in G, and through G draw
EGF touching the circle^ and meeting DA produced in E, and DB

K tt I*

produced in F ; "EF will be the side of an equilateral polygon de<
8cri6ed about the circle ABC (3. 1. Sup.)* In AC produced take
AK equal to half the perimeter of the polygon whose side is AB $
and AL equal to half the perimeter of the polygon whose side is £F<
Then AK will be less, and AL greater than the stritight line AH
(Ax. 1. Sup.). Now, because in the triangle EDF, DG is drawn
perpendicular to the base, the triangle EDF is equal to the rectangle
contained by DG and the half of £F (41. 1.) ; and as the same is true
of all the other equal triangles having their vertices in D, which make
up the polygon described about the circle ; therefore, the whole poly.
goa is equal to the rectangle contained by DG and AL, half the peri-
meter of the polygon (1. 2.), or by DA and AL. But AL is greater
than AH» therefore the rectangle DA . AL is greater than the rectangle
DA. AH ; the rectangle DA. AH is therefore less than the rectangle
DA.AL, that is, than any polygon described about the circle ABC.
Again, the triangle ADB^s equal to the rectangle contained by DM
the perpendicular, and one hidf of the base AB, and it is therefore
less than the rectangle contained by.DG, or DA, and the half of AB#

• K H I.

OF GEOMETRY. QOOK L 160

And as the Bame is true of all the other triangles having their verti*
ces iu D, which make up the inscribed polygon, therefore the whole
of the inscribed polygbn is less than the rectangle contained by DA»
and AK half the perimeter of tfie polygon.' Now, the rectangle
DA.AK is less than DA.AH ; much more, therefore, is the polygon
whose side is AB'less than DA.AB ; and the rectangle DA.AH is
therefore greater than any polygon inscribed in the circle ABC. But
the same rectangle DA.AH has been proved to be less than any poly*
^on described about the circle ABC ; therefore the rectangle DA.AH
is equal to the cirtsle AB€, (2. Cor. 4. 1. Sup.) *Now DA -is the se-
midiameter of the icircle ABC, and AH the half of its circumference*
Therefore, &c. a. E. D.

CoR. 1. Because DA : AH : :: DA^ ^ DA,AH (1. 6.), and because
by this proposition, DA.AH== the area of the circle, of which DA is
the radius : therefore, as the. radius of any circle to the semicircunv-
ference^ or as the diameter to the whole circumference, so is the
square of the radius to the area of the circle.

CoK.^. Hence a polygon may be described about ja circle, thepe*
rimeter of which shall exceed the circumference of the xnrcle by a
line that is less than any given line. Let NO be the given line.
Take in NO the part NP less than its half, and also than AD, and let
a polygon be tSescribed about the circle ABC, so that its excess above
ABC -may be less than the square of NP, (I. Cor. 4. 1. Sup.)* Let •
the side of this polygon be EF. And since, as has been proved, the •
circle is equal to the rectangle DA.AH, anfd the polygon to the rect-
angle DA; A-L, the excess of the polygon above the circle ts equal to
the rectangle DA.HL ; therefore the rectangle DA.HL is less than
the square of NP ; and therefore, since DA is greater than NP, HL' .
is less than NP, and twice HL less than twice NP, wherefore, much
more is twice HL less than NO. But HL is the difference between •
half the perinxeter of the polygon whose side is EF, and half the cir-
cumference of the circle ; 'therefore, twice HL is the difference be-
tween the whole perimeter of the polygon and the w^ole circumfer-
ence of the circle (5; 6.).. The difference, therefore, between the
perimeter of the polygon and the circumference of the circle is less
than the given line NO.

Cor. 3. Hence also, a polygon may be inscribed in a circle, such
that the excess of the circumference above the perimeter of the poly-
gon may be less than any, given line. This is proved Uke the pre^

ceding.

PROP. VL THEOR-

l!%e areas sf circles are to one another in the dupUcaie rathj or as lh»

squaresqftiteir diameters.

Let ABD and GHL be two circles, of which the diameters are AD
and GL ; Ae circle ABD is to the circle GHL as the square of AD

la the square of <3L*

170

SUPPLEMENT TO THE ELEMENTS

Let ABCDEF and GHKLM N be two equilateral polygona of tber
same number of sides inscribed in Che circles ABD, GHL; and let^

be such a space that the square of AD is to the square of GL as the'
circle ABD to the space Q. Because the, polygons ABCDEF and
GHKLMN are equilateral and of the same number of sides, they
are'similar (2. 1. Sup.)> and their areas are as the ^squares of the
diameters of the circles in which they are inscribed. Therefore
AD» iQUi: polygon ABCDEF : polygon GHKLMN ; but AD« :
GL" : : 6irc|e ABD : Q ; and therefore, ABCDEF : GHKLM : :
circle ABD s Q. Now, circle ABD 7 ABCDEF ; therefore Q 7
GHKLMN (14. 5»), that is, d is greater than any polygon inscribed
in the circle GHL.

, In the same manner it is demonstrated, that Q is less than any pO'
lygon described about the circle GHL ; wberefore the space Q is
equal to the circle GHL (2. Cor. 4. 1. Sup.). Now, by hypothesis^
the circle ABD is to the space Q as the square of AD to the square
of GL ; therefore the circle ABD is to the circle GHL as the square
of AD to the square of GL. Therefore, &c. Q. E. D.

Cor. 1. Hence the circumferences of circles are to one another
as their diameters.

OP GEOMETRY. BOOK L 171

Let the straight line X be eqvial to half the circumference of the
circle ABD, and the straight line Y to; half the circumference of the

circle GHL : And because the rectangles AO.X and GP.Y are equal
to the Circles ABD and GHL (5. 1. Sup.) therefore AO.X : GP.Y
: : AD^ : GL^ : : AO' : GP* ; and alternately, AO.X ; AC|^ : : GP,Y
: GP'' ; whence, because rectangles that have equal altitudes are as
theif bases (1. 6.), X : AO : : Y : GP, and again alternately, X : Y
: : AO : GP : wherefore, taking the doubles of each, the circumfer-
ence ABD is to the circumference GHL as the diameter AD to the
diameter GL* * ' .

Cor. 2. The circle that is described upon the side of a right an-
gled triangle opposite to the right angle, is equal to the two circles
'described on the other two sides. For the circle described upon SR
is to the circle described upon RT as the square of SR to the square
of RT ; and the crrcle -described upon TS is to the circle described
upon RT as the square of ST to the square of RT. Wherefore,
the circles described on SR and
on ST are to the circle describ-
ed on RT as the squares of 8R
and of ST to the square of RT
(24. 5.).. But the squares of RS
and of ST are equal to the
square of RT (47, 1.) ; there-
fore the circles described on RS
and ST are equal to the circle
described on RT

PROP. Vir. THEOR.

Equiangular parallelograms are to one another as the products of the

numbers proportional to their sides.

Let AC and DF be two equiangular parallelograms, and let M, N,
P and Q be four numbers, suoh that AB : BC : : M : N ; AB : DE : :
M : P ; and AB : EF : : M : Q, and therefore ex sequali, BC : EF : :
N : Q. The parallelogram ACis to the parallelogram DF as MN
to PQ. •

Let NP be the product of N into P, and the ratio of MN to PQ
will be compounded of the ratios (def. 10. 5.) of MN to NP, and of
NP to PQ. But the ratio of MN to NP is the same with that of M

' 1

172

SUPPLEMENT TO THE ELEMENTS

B D

to P (15. 5.), because MN and NF are equimultiples of M and P^
and for the same reason, the ratio of NP to PQ is the same with that
of Njo Q ; therefore the ratio of MN to PQ is compounded of the
ratios of M to P^ and of N to Q. Now, the ratio of M to P is the
same with that of the side AB to the side DE (by Hyp.) ; and the
ratio of N to Q the same with that of the side BC to the side £F»
Therefore, the ratio of MN to PQ is compounded lof the ratios of
AB to DE, and of BG to EF. And the ratio of the parallelogram AG
to the parallelogram DF is compounded of the same ratios (23. 6.) ;
therefore, the parallelogram AG is to the parallelogram DF as MN,
the product of the numbers M and'N, to PQ, the product of the num»
bers P and Q. Therefore, ISlc. Q. E. D.

Cor. I. Hence, if GH be to KL as the number M to the number

N ; the square described on GH • '

will be to the square described on G H K ~ L*

KLas MM, the square of the num-
ber M to NN ; the square of the number Nr

Cor. 2. If A, )3, C, D, &c. are any lines, and m, », r, «, &c. num>''
bers proportional to them ; viz. A : B : : m : n, A : G : : m : r, A :.
D: im : 8, &c. ; and if the rectangle contained by any two of the
lines be equal to the square of a third linej the product of the num-
bers proportional to the first two, will be equal to the square of the
number proportional to the third ] that is, if A.C=B^ mXr^nXfiy
or =n*.

For by this Prop, A.C : B' : : mYr :ri^; bat A.C=BS therefore
wrXr=^n'. Nearly in the same way it may be demonstrated, that
, whatever is the relation between the rectangles contained by these
lines, there is the same -between the products of the numbers propor-
tional to them.

'• So also conversely if m and r be numbers proportional to the linee
A and C ; if also A.C=B', and if a number n be found such, that n*
=imr, then A : B : : m : n. For let A : B : : m : g, then since m, q^
r are proportiohal to A, B, and C, and A.G=B^; therefore, as has
just been proved,. 5''=mXr ; but n'—'^'Xr, by hypothesis, therefore
v^^^y and n=s=5' ; wherefore A : B : : m : n.

SCHOLIUM.

la order io have numbers proportional to any set of magnitudes of
the same kind, suppose one of them to be divided into any number, m
of equal parts, and let H be one of those parts. Let H be found u

OP 6E0METRT. BOOK I.

17S

times in the magnitude B, r times in C, s times in D, ^c, then it is
evident that the numbers m,.n} r, « are proportional to the magnitudes
A, B, C and D. When therefore it is saidln any of the following pro-
positions, that a line as A= a number m, it is understood that A^m
XH» or that A is equal to the given magnitude H multiplied by m,
and the same is understood of the other magnitudes, B, C, D, and
their proportional numbers, H being the common measure of all the ,
magnitudes. This common measure is omitted for the sake of "bre-
vity in the arithmetical expression ; but is always implied, when a
line^ or othergedmetrical magnitude, is said to be equal to a. number.
Also, when there are fractions in the number to* which the magnitude
is called equal, it is meant that the common measure H is farther sub.
divided into such parts as the numerical fraction indicates. Thus, if
As:360..375, it is meant that there is a certain magnitude H9 such that

A=360XH+Y-— XH, or that A is equal to 360 times H, together

with 375 of the thousandth parts of H. And the same is true in all
other cases, where numbers are used to express the relations of geo*
metrical magnitudes.

PROP. VIII. THEOR.

The perpendicular drawn from the centre of a circle on the chord of any
arch is a mean proportional between half the radius and the Une made
up of the radius and the perpendicular drawn from the centre on the
chord of double that arch : ,Snd the chord of the arch is a mean pro-
portional between the diameter and a line; which is the difference be-
tween the radius and the aforesaid perpendicular from the centre.

Let ADB be a circle, of which the centre is C 5 DEE any arch,
and DB the half of it ; let the chords DE, DB be drawn : as also CF
and CG at right angles to DE and DB ; if OF be produced it will
meet the circumference in B : let it meet it again in A, ^nd let AC be
bisected in H; CG is a mean proportional between AH and AF; and

174

SUPPLEMENT TO THE ELEMENTS

BD a mean proportional between AB and BF,the excess of the ra-
dius above CF. '

Join AD ; and because ADB is a right angle, being an angle in a
semicircle ; and because CGB is also a right angle,, the triangles
ADD, CBG are equiangular, and, AB : AD : : BC ; CG (4. 6.), or
alternately, AB : BC : : AD : CG ; and therefore, because- A B is
double of BC, AD is double of CG, and the square of AD therefore
equal to four times the square of GG«

> But, because ADB is a right angled triangle, and DP a perpendicu-
lar on AB, AD is a mean proportional between AB and* AF (8. 6.)^
and AD'=AB.AF (17. 6.), or since AB is =4AH, AD^=^4AH.AF.
Therefore also, because 4CG2=:AD^ 4CG2=4AH. AF, and-eG2=
AH. AF ; wherefore CG is a mean proportional between AH and AF,
that is, between half the radius and the line made up of the radius,
and the perpendicular on the chord of twice the arch BD.

Again, it isr evident that BD is a mean proportional between AB
and BP (8. 6.), thnt is, between the diameter and the excess of the
radius above the perpendicular, on the chord of twice the arch DB.
Therefore, &c, Q. E. D.

PROP. IX. THEOR.*

The circumference of a circle exceeds three iirrij^ thediamefery by a line
less than ten of the pariSy of which the diameter contains seventy, hit
^greater than ten of the parts whereof the diameter contains seventy -one.

Let ABD be a circle, of which the centre is^C, and the diameter
AB \ the circumference is greater than three times AB, by & line less

tha!n — , or -, of AC, but greater than — of AC.

In the circle ABD apply the straight line BD equal to the radius

» lu this proposition, (he character + placed after a number, signifies thai something is
Jo be added to it ; and the character — , on the other band, signifies th

^cn awa^ from it,

that something^ is to foe

OP GEOMETRY. BOOK I, IW

be : Draw DF perpendicular to BO, and let it meet the circuinf<»«
ence again in E ; draw also CG pcrdendiculat to 3£h produce BG
to A, bisect AC in H, and join CD.

It is evident, that the arches BD, BE are each of them one-sixth
of the circumference '(Cor. 15. 4.), and that therefore the areh DB£
is one third of the circumference. Wherefore, the line (8. 1. Sup.)
CG is a mean proportional between All, half the radius, and the line
AF« Now because the sides BD, DC, of the triangle BDC are equa),
the angles DCF, DBF are also equal ; and the tingles DFC, DFB
being equal, and the side DF common to the triangles DBF, DCF*
the base BF is equal to the base CF, and BC is bisected iq F. -

Therefore, if AC or BC=1000, AH=:500, CF=500, AF=1500,
and CG being a mean proportional between AH and AF, CG^e3(17»
6.) AH. AF=600X 1500=750000; wherefore CG= 866.0254+ ,
because (866.0254)^ is less than 750000. Hence also, AC+CG«s:
1866.0254+.

Now, as CG is the perpendicular drawq from the centre Cf on the
chord of one-sixth of the circumference, if P = the perpendicular
from C on the chord of one-twelflh of the circumference, F will be a
mean proportional between AH (8. 1. Sup.) and AC+CG, and P=
AH (AC+CG)=500 X (1866.0254+) =933012.7+. Therefore,
F^965.9258+, because (965.9258)' is less than 933012.7. • Hence
also, AC+P=1965.9258+.

Again, if Q= the perpendicular drawn from C on the chord of one
twenty-fourth of the circumference, Q will be a mean proportional
between AH and AC+P, and Q'=AH (AC+P)=600(1965.9258+)
=982962.9+ ; and therefore Q=991.4449+, because (991.4449);*
is less than 982962.9; Therefore also AC+a==1991.4449+.

In like manner, if S be the perpen4icular from C on the chord <of
one forty-eighth of the circumference,' S' ="AH (AC+Q) = 500
(1991.4449+.)=995722.45+ ; and S=997.8589+, because (997.
8589)2 i^ legg than 995722.45. Hence also, AC+S =1997.8589+,

Lastly, if T be the perpendicular from C on the chord of one nine-
ty-sixth of the circumference, T^ = AH (AC+S)=500 (1997.858^
+) = 998929.45+, and T = 999.46458 +. Thus T, the perpen^
dicular on the chord of one ninety-sixth of the circumference, is
greater than 999.46458 of those parts of which the radius contains

1000.

But by the last proposition, the chord of one hinety-sixth part of '
the circumference is a mean proportional between the diameter and,
the. excess of the radius above S, the perpendicular from the centre
on the chord of one forty-eighth of the circumference. Therefore»N
the square of the chord of o4e ninety -sixth of the circumference ==>
AB ( AC— S) = 2000X(2.1411— ,) = 4282.2— ; and therefore the
chord itself=65.4386^, because (65.4386)' is greater than 4282.2,
Now, the chord of one ninety*sixth of the circumference, or the side
of an equilateral polygon of tiinety.six sides inscribed in the cix<!le.

v-^

lie

SUPPLEMENT TO THE ELEMENTS

being 65.4396 — , the perimeter of that polygon will be s= (65,4386
.) 96 = 6282.1056—.

Let the perimeter of the circumscribed polygon of the same num-
ber of sides^be M, then (2, Cor, 2. 1. Sup.) T : AC : ; 6282.1056—
: M, that is, (since T=999.46458+, as already shewn),

099.46458 + : 1000 : : 6282.1056— : M ; if then N be
«uch,"that 999.46458 : 1000 : : 6282.1056— : N ; ex aequo per-
turb. 999.46458 + : 999.46458 : : N : M ; and, since the first ifiu
greater than the second, the third is greater than the fourth, or N is
greater than M.

Now, if a fourth proportional be found to 999.46458, 1000 and
6282.1056 viz. 6285.461—, then,

because, 999.46458 : 1000 : : 6282.1056 : 6285.461--,
and as before, 999.46458 : 1000 : : 6282.1056— : N ;
therefore, 6282.1056 : 6282.1056— : : 6285.461— N, and as the first
o{ these proportionals is greater than the second, the third, viz. 6285.

461 — is greater than N, the fourth. But N was proved to he greater
than M ;. mUch more, therefore,, is 6285.461 greater than M, the peri*
meter of a polygon of ninety-six sides circumscribed about the circle ;
that is, the perimeter of that polygon is less than 6285.461 ; now,
the circumference of the circle is less than the perimeter of the po-
^lygon ; much more, therefore, is it less thai^- 6285.461 ; wherefore
the circumference of a circle is less than 6285.461 of those parts of
which the radius contains 1000. The circumference,*lherefore, has
to the diameter a less ratio (8. 5.) than 6285.461 has to 2000, or than
3142.7305 has to 1000 : but the ratio of 22 to 7 is greater than jtbe
ratio of 3142.7305 to 1000, therefore the circumference has a less
ratio to the diameter than 22 has to 7, or the circumference is less
*tban 32 of- the parts of which the diameter contains 7.

&i^ 6£OM£lrRY. Bookii irr

It remains to demonstrate, that the pairt by ^ich the circumference
exceed^ the diameter is greater than ^ of the diameter. '^

It was before shewn, that C;G*==:*760000 ; wherefore Ci:?^^^^^
b2545— , because (866,0:3545)' is grisater than 750000$ dierefore
AC+CX>= 1866.02645— .

Now» P being, as before, ^he pi^rpehdicular from the centrie oki
the chord of one twelfth of the circumference; P*==AH (AC+CG)
»500x(i866,«$2545)— r=933012.'73^; and P=965.92585~, be-
4Bause (965^2585)' is greater than 63301 2.73; Hence also, AC-f|^
iBl965.92585_i

Next, as Qs=.the perpendicular drawn ftotn the centre on the chord
of one twenty-fourth of the circumference, Q'==» AH (AC^rP) =*
600 X (1965.92585— ) =982962.93— J and Q = 991.44495-^ be-
cause (991.44495)'' is greater than 982962.93. Hence alsoj At)+Q
2=1991.44495^; . ,

In like manner, as 8 is the perpendicular ftcm O on ^e tboi^ of
♦)ne forty. eighth of the cirbumference^ B"=AH (AC+Q)— 500(1991;
44495_) = 995722.475—, and S = (997.85865—) becain^ (9*7;
.65895)' is greater than ^722.475. ,

But the square of the diord of the ninety-sixth part of the circum-
fcfrence = AB (AC— S) = 2000 <2.i4105+) .= 4282.1 -f, and thd
^hord itself ==: 65.43t7 + because (65.4377)' is less than 4282,1 :
Now the chord of one ninmy-sixth part of thie circumference being
^=65.4377+j the perimeter of a polygon of ninety-six sides inscribe
kd in the fcircle=(65i437'i'+)96 =6282 .019+. But the circumfer-
ence of the circle is greater than th^ perimeter of th^ inscfibbd.
polygon ; therefor^ the circumference is greater than 6282.019^ oC
those parts of which the radius contains 1000; or than 3141.009 of
thh parts of which the radius contaitid 500j or the diameter contains

'^ * 10 '

iOOO. Now, 3141(00^ has to 1000 a greater ratio than 3+^ to 1 ;

therefore thil circumference of the circle has a greater ratio to thd

• 10
diameter than 3+;^ has to 1 ; that is, the excesd of the circumference

Hbove three time^ the diameter is gteatei' than ten of itiose parts o^
li^hich the diameter contains 71 ; and it has already been sheWn to be
less than ten of those of which the diameter Contains 70« Therefore^
Ac. Q, E. D,

CoH. 1 ; Hence th& diameter of a circle being given, the circumfer*
^nce may be found nearly, by making as 7 16 22, so the givbn diame.
ter to a foiiHh pYoportionaT, which will be greater than the circumfer-

(bnce. And if as 1 to 3HK=y» or as 71 or 22$, so the given diameter

to a fourth proportional, this will be nearly equal t6 the cironmfer'
^ncei but will be less than it.

Its

SUPPLEMENt TO THE ELEWENTSr

Cob. 1. Because the difference b)9tween - and ^^r- is

7 71

497'

th^re*^

fore the Unes found by these proportionals di&er by j^of the diame--

ter. Therefore the difference ef either of them from the circumfer*
ence must be less than the 497th part of the diameter;

Cos. 3. As 7 to 22, so the square of the radius to the area- of the
circle nearly.

For it has been shewn, that (1. Cor. 5. 1. Sup.) the diameter of a
circle is to its circumference as the square of the radius to the area
of the circle ; but the diameter is to the circumference nearly as 7 ti»
22, therefore the square of the radius is to the area of the circle near-
ly in that same ratio.

SCHOLIUM.

- It is evident that the method employed in thi$ proposition, for find-
ing the limits of the ratio of the circumference of the diameter, may
be carried' to a greater degree of exactness, by finding the perimeter
of an inscribed and of a circumscribed polygon of a greater number
ef sides than 96. The manner in which^the perimeters of such po-
lygons approach nearer to one another, as the number of their sides
increases, may be seen from the following Table, which is construct-
ed oh the principles explained in the foregoing Proposition, and iO'
which the radius is supposed =- 1.

NO. of Sides

Perimeter of the

Perimeter of t&e

of the Poly-

inscribed Poly-

circumscribed

gon.

Polygon.

6.000000

6.822033

6.211657+

6.430781—

24-

6.265257+

6.319320—

6.278700+

6.292173

6.282063+

6.285430—

192

6.282904+

Q.283747—

384

6.283116+

; 6.283327—

768

6.283167+

6.283221—

1536

6.283180+

6.283195—

3072

6.283184+

6.283188

6144

6.283185+

6.283186—

The part that is wanting in the numbers of the second column, td*
make up the- entire perimeter of any of the inscribed polygons, is^
less than unit in the sixth decimal place; and in like manner, the
part by which the numbers in the last column exceed the perimeter
of any of the circumscribed polygons is less than aunit in the sixtlv^

OFGEOMETRt. BOOKL

179

^decimal place, that is, than

of the radius. Also, as the num-

1000000

bers in the second column are less than the perimeters of the ii^.
«cribed polygons, they are each of them less than the circumference
of the circle ; and for the same reason, each of those in the third
-column is greater than the circumference. But when the arch of

>r- of the circumference is bisected ten times, the number of sides in
the polygon is 6144, and the numbers in the Table differ from one
another only by ^ r— part of the radius, and therefore the peri-
meters of the polygons differ by less than that quantity ; and conse-
<|uently the circumference of the circle, which is greater than the
least, and less than the greatest of these numbers, is determined with*
In less than the millioneth part of the radius.

Hence also, if R be the radius of any circle, the circumference is
greater than R X 6.283185, or than 2RX 3. 141592, but less than 2R
X 3.141 593 ; and these numbers differ from one another only by a
millioneth part of the radius. So also R^+3.141592 is less, land R'
X 3. 141 593 greater than the area of the circle ; and these numbers
differ from one another only by a millioneth part of the square of the
radius.

In this way, also, the circumference and the area of the circle
may be found still nearer to the truth ; but neither by this, nor by
any other method yet known to geometers, can they be exactly de-
Jtemained, though the errors of both may be reduced to a less quantity
Ihan any that can be assigned.

EI^EMENTS

GEOMETRY,

SUPPLEMENT.

BOOK il,

<«P THE INTERSECTION OF PLANEa

DEJ^NITIONS.

A 9TKxiGWB niie is perpendicular or at rfgfit angles fa a pHaiie^ -wfkem
it makes right angles wi^h ^very straigbt Hne which it meets ih^
ibat planer

^ plane is perpendicular io 9. plane, when the straight line^ drawQ^
m one of the planes perpendicular to the common section of th»<
two planes, are perpeodicuiajc to the oUiqk plane.

IIL.

Xhe inclination of a sti^ight line to a plane is the aeute* angle con>..
lained by that straight line, and another drawn froQi the point in^
which the first Hne meets the plane, to the point in which a per-i
pendicular to the plane^ drawn, fiom any point of t^e firal fin^,,
meets the sanoie plane.

IV.

Ifhe angle made by two planes which cut one another, is the angle^
contained by two straight lines drawn from any, the same point iq^
the line of their common section, at right angles to that line, the
one, in the one plane, and the other, in the othes. • Of the two ad*,
facent angles made by two lines drawn in this manner, that whicl\
^eiCUte i9 «i^. QaU^d, tbie io^clii^i^tiw of the planes to OQe anothei^^

SUPPX.BB4ENT TO THE ELEMENTS, He. 181

Two pHpes are stiid to have the same, or a like inclriiQtioh to one aii**
other, which two oth^r planes have, when the angles of incliniifioti
above defined are equal to one anotlier.

VI.

A straight line is said to be parallel to a plane, when it does not
meet the plane, though produced ever ao far*

VII.

PTanes are i^aid to be pi^rallel td dnw enbthenr, wbieh do not meet,
though produced oveir so fan

vni.

A solid angle is an angle made by the meeting of more than two
' ' plane angles, which are not i^ the same plan^ in one point.

PROP. L THEOR.

One pari of a straight line cannot he in d plaHe and another pari about ft.

If itbe possible let AB,part6f t^b straight Itnd ABC, be in the plane^
^iid the part BC above it : and since
the straight line A B is in the plane, it
can be produced in that plane. (2. Post.
1.) ; let it be produced to D : Then
ABC and A BD arc two straight lines,
^nd they Kave the common segment
AB, which'is impossible (Cor. dief. 8.
1.) Therefore AQC is nqt a straight IJne. Wherefore one part, *o,

ft? E?p.

PROP. II. THEOR;

^.ny three straight Unes which iheet drts anathen, not in the samepoini^

are iti one plane*

Let the three straight Hues AB, CD, CB ni^et dhe another in the
points B, C and E ; AB, CD, OB are in one
plane.

Let any plane pass through the straight
line EB, and let the plane be turned about
EB, produced, if necessary, until it pass
^hrQugh the point C : Then, because the
points E, C are in this plaho, the straight
line EC is in it (def. 5. 1.) : for the same
reason, the straight line BC isjii the sanie ;
and, by the hypothesis, EB is in it ; there-
fore the three straight lines EC, CB, BE are
}n c^ne plaiie ; but the wliole of the lines DC,

183 SUPPLEMENT TO THE El^MENTS

AB, and BC produced, are ia the same plane with th% parts of them
EC, EB, BC (1. 2, Sup.) Therefore AB, CD, CB, are all ia on«
plane. Wherefore, <&o. Q. E. D.

Cob. It is manifest, that any two straight lines which cut one an-
other are tn one plane : Also, that any three points whatever are in
one plane.

PROP. nr. THEOR.

ijr tm jianea ad one another, their common section is a strmght line.

Let two planes AB, BC cut one ano-
ther, and let B and D be two points in the
line of their common section. From B to
D draw the straight line BD ; and because
the points B and D are in the plane AB,
the straight line BD is in thai plane (def. ri'^

5. 1.) : for the same reason it is in the '•'^yj
plane CB ; the straight line BD is there. j^

fore common to the pjaoea AB and BC, or
it ia the commoa section of these planes. Therefore, &c. Q. E. D.

PROP. IV. THEOR.

If a ttraighi line stand at right anglex to each of two straight Itaes in
the point of their intersection, it wUl alto be at right angles to the
plane in which these linesare.

Let the strai^t line AB stand at right angles to each of the
straight lines BF, CD in A, the point of their intersection : AB is
also at right angles to the plane passing through EF, CD.

Through A draw any lino AG in the
plane in which are EF and CD ; let G
be any point in that line ; draw GH pa-
rallel to AD ; and make HF=HA, join
FG ; and when produced let it meet
CAinD; joinBD.BG, BF. Because
GH is parallel to AD, and FH=HA :
therefore FG=GD, so that the line DF
is bisected in G. And because BAD is
aright angle,BD»=AB*+AD'(47. 1.) ;
and for the same reason, BF"=AB'+
A^, therefore BD' + BF' = 2AB' +
AD'-HAF' i and because DF is bisect,
ed in G (A. 2.), AD'-i-AF==2AG'4-

aGF", therefore BD'-(-BF'=2AB"-|-2AG'-|-2GP. But BD"-;-
3F»=>(A. 3.) 2BG»+2GF', therefore SBG" + 2GP=2Aff -(.

OP GEOMETRY. BOOKII.

183

4AG*+2GP; and taking 2GP from both, 2BG^2AB«+2AG*, or
BG'=AB«+AG» ; when BAG (48. 1.) is a right angle. Now AG
i^ any straight lino drawn in the plane of the lines AD, AF ;. and
when a straight lihe is at right angles to any straight line which it
meets with in a plane, it is at right angles to the plane itself (def. I.
S. Sup.). AB is therefore at right angles to the plane of the lines
AF, AD. Therefore, dec. Q. E. D.

PROP. V. THEOR.

if three glraighi lines meet dt in one pointy and a straight line ttandai
right angles to each of them in that point ; these three straight lines
are in one and the same plane.

Let tfacr straight Hne AB stand at right angles to each of the straight
fines BC, BD, BE, in B, the point where they meet ; BC, BD, BE
are in one and the same plane.

' If not, let BD and BE, if possible, be in one plane, and BC be
above it ; and let a plane pass through AB, BC, the common section
of which with the plane, in which BD and BE are, shall be a straight
73. 2. Sup.) line ; let this be BF : therefore the three cflraight lines
AB, BC, BF are all in one plane, viz. that which passes through AD,
BC ; and because AB stands at right angles to each of the straight lines
BD, BE, It is also i^t right angles (4. 2. Sup.) to^th^ plane passing
through them ; and therefore makes right J^
angles with every straight line meeting it in
that plane ; but BF which is in that plane
meets it; therefore the angle ABF is a right
angle ; but the angle ABC, by the hypothe-
sis is also a right angle ; therefore the an-
gle ABFis equal to the angle ABC,and they
are both in the same plane, yrhich is impos-
sible : therefore the straight line BG is not jq
above the plane in which are BD and BE :
Wherefore the three straight lines BG, BD,

BE are in one and the same plane. There. ^^K

lore, if thMe straight lines, &c. Q. E. D.

PROP. VI. THEOR.

Two straight lines which are at right angles to ike same ptane, af^

parallel to one another.

Let the straight lines AB, CD bo alright angles to the same plane
BDE ^ AB is parallel to CD. .

184

StTPPLllMENT TO ^£tE UtEMlli^f g(

^^ t

Let them nieet the plane in the points B, D. Draw.DE at I'ight
Imgles to DBy in the plane Bl>£, and let £ be A w

any poiilt in it : Join AE, AD, EB. Beeatnrcf ^ ^

ABE is a fight angle, AB' + BE' « (47. L)
AE', and because BDE is a right angle, BB^^ ^
BD« + DE» ; therefore AB» + Siy + rffiy^
AE»; now, AB*+BD*^ AD", because ABD is
a right angle, therefore AD'+D£'=AE', and
ADE is therefore a (48. 1.) right angle. There- B
fore ED is perpendicular to the three lines BD,
DA, DCi whence these lines are in one plahlei
(5. 2. Sup.)< Bert AB is in the plane in which
are BD, DA, because any three straight lines,
#hieh meet one another, are in one plane (2..12. S

8up.) : therefore AB, BD, DC are in one plane ; ancf each of thcr
angles ABD, BDC is a right angle ; therefore AB is parallel (28^.^
1.) to CD. Wherefore if two straight lines, dsc. Q. £. D.

PROP. VII. THEOR.

Jifiwo straight lines he par allele arid one of (hem al tighi angles id a
plane ; the other is (dso at right angles to the same plane.

G-

Let AB, CD be two parallel straight lines', atid let one of them ASf
he at right angles tor a plane ; the a f%

other CD is at right angles to the -^ ^

same plane.

For, if CD be not perpendicU-
far to the plane to which AB is per-
pendicular, let DQ be perpendiciT.
kr to it. then (6. 2. Sup.) DG is
parallel to AB : DG add DC thet€* E,
lore are both parallel to AB, and
are drawn through the same point
t), which is impossible (II. Ax. I.)
Therefore, &c. Q. E. D.

JMi.

PROP. Vm. tHEOR/

Two straight lines which are each of them peiraUel to the same straight
line, though fiat both in the s^ame j^nieti^ith d, Ore parallel to one ari*
either*

Let AB, CD be each of them paralM to E^, and rtot th the same
plane with it ; AB shall be parallel to CD.

tn EF take any point G, from which draw, in the plane passing
through £F, AB, the straight line GH at right angles to EF; and ki'
the plane passing through EF, CD^ draw GK at right angles to \h^

DP GEOMfiTRT. BOOK 11.

11BS

saYHe Et". And because EF is perpendicular both to GH and GK, it
is perpendicular (4. 2. Sup.) to the prane HGK passing through
them : and EF is parallel to AB ; therefore AB is at right angles
<7. 2. Sup.) to the plane HGK.

For the same reason, CD is J^ 5— — B

likewise at right angles ta th«
plane HGK. Therefore AB,
CD are each of them at right
angles to the plane HGK. But
if two straight lines are at right
angles to the same plane, they
are parallel (6. 2. Sup.) to one G
another. Therefore AB is pa-
rallel to CD. Wherefore two straight lines, &c. Q. E. D.

PROP. IX. THEOR.

If two straight lines meeting one another he parallel to two others thai
meet one_ another, though not in the stone plane with the first two ; fAtf
jUfst two and the other two shaU contain equal angles.

Let the two straight lines AB, BC which meet one another, be pa*
rallel to the two straight lines DE, EF that meet one another, and
are not in the same plane with AB, BO. The _

angle ABC is equal to the angle DEF. -"

Take BA, BC, ED, EF all equal to one
another ; and join AD, CF, BE, AC, DF : j.
Because BA is equal and parallel to ED, there^*^
fore AD is ^33. 1.) both equal and parallel to
BE. For the same reason, CF is equal and
parallel to BE. ^Therefore AD and CF are
each of them eqiial and parallel to BE. But
straight lines that are parallel to the same
straight line, though not in the same plane
with it, are parallel (8. 2. Sup.) to one ano-
ther. Therefore AD is paraltel to CF ; and
it is equal to it, and AC, DF join them towards D'
the same parts ; and therefore (33. 1.) AC is equal lind parallel to
DF. And because AB, BC are equal to DE, EF, and the base AC
to the base DF^; the angle ABC is equal (8. 1.) to the angle DEF.
Therefore, if two straight lines, &c. Q. E. D.

PROP. X. PROB.

To draw m straight line perpendicular to a pUme^from a given point

above it*

Let A be the given point above the plane BH, it Is required to draw
from the point A a straight line perpendicular lo the plane BH«

186 SUPPLEMENT TO THE ELEMENTS

In the plane draw any straight line BC, and from the point A draw
(12. 1.) AD perpendicular to BC. If then AD be also perpendicu-
lar to the plane BH, the thing required is already done ; but if k
be not, from the point D draw
(11. 1.), in the plane BH, the
straight line DE at right angles to
BC ; and from the point A draw
AF perpendicular to DE; and
through F draw (31. 1.) QH pa-
rallel to BC : and because BC is
at right angles to ED, and DA,
BC is at right angles (4. 2. Sup.)
to the plane passing through ED,
DA. And GH is parallel to BC ; but if two straight lines be parallel,
one of which is at right angles to a plane, the other shall be at right
(7. 2. Sup.) angles to the same plane ; wherefore GH is at right an*
gles to the plane through ED, DA, and is perpendicular (deC I. 2*
Sup.) to every straight line meeting it in that plane. But AF, which
is in the plane through ED, DA, meets it : Therefore GH is perpen-
dicular to AF, and consequently AF is perpendicular to GH ; and
\F is also perpendicular to DE : Therefore AF is perpendicular to-
each of the straight lines GH, DEL But if a straight line stands at
right angles to each of two straight lines in the point of their inter,
section, it is also at right angles to the plane passing through then}
(4. 2. Sup.). And the plane passing through ED, GH is the plane
BH ; therefore AF is perpeodicalar to the plane BH ; so that, front
the-given point A, above the plane BH, th6 straight line AF is drawn
perpendicular to that plane. Which was to be ckHte.

OoR. If it be required from a point C in a plane to erect a per-
pendicular to that plane, take a point A above the plane^ and draw
AF perpendicular to the plane ; then, if from C a line be drawn pa-
rallel to AF, it will be the perpendicular required ; for being paral*
le\ to AF it will be perpendicular to the same plane to which AF is^
perpendicular (7. 2. Sup.).

PROP. XI. THEOR.

From the same poisit in a plane, there cannot he ttoo straight lines at
right angles to the plane, upon the same side of it; And there can he
' h%U one perpendicular to a plane from a point above it.

For if it be possible, let the two straight lines AC, A B be at right
angles to a given plane from the same point A in the plane, and upoa
the same side of it ; and let a plane pass through BA, AC ; the con[N
mon section of this plane with the given plane is a straight (3. 2. Sup.)
line passing through A : Let DAE be their common section : There-^
fore the straight lines AB, AC, DAE are in one plane : And because
CA is at right angles ta the given plane^ it makes right aagles.witb

OP GEOMETRY. BOOK 11.

187

«vei7 straight line meeting it in that plane.
But DAE, which is in that plane, meets
CA : therefore CAE is a right angle. For
the same reason BAE is a right angle.
Wherefore the angle CAE is equal to the
angle BAE ; and they are in one plane,
which is impossible. Also, from a point

above a plane, there can be but one per- ^ ^ — -

pendicular to that plane ; for if there could ^ ^ K

be two, they would be parallel (6. 2. Sup.) to one another, which is
absurd. Therefore, from the same point, &c^ Q. E. D.

PROP, XII. THEOR.

Planes to which the same straight line is perpendicular, are parallel to

one another*

Let the straight line AB be perpendicular to each of the planes CD,
EF : th^se planes are parallel^ to one another.

If not, they must meet one another when produced, and their com-
mon section must be a straight line 6H, in
which take any point K, and join AK, BK :
Then, because AB is perpendicular to the
plane EF, it is perpendicular (def. 1. 2. Sup.)
to the straight line BK which is in that plane,
and therefore ABK is a right angle. For the
same reason, BAK is a right angle ; where-
fore the two angles. ABK, BAK of the trian-
gle ABK are equal to two right angles, which
is impossible, (17. 1.) : Therefore the planes
CD, EF, though produced, do not meet one
another ; that is, they are parallel (def. 7. 2.
Sup.}. Therefore planes, <&c. Q. E. D.

PROP. XIII. THEOR.

if tioo straight lines meeting one another, heparaUd to two straight lines
which also meet one another, hut are not in the same plane with the
first two : the plane which passes through the first two is parallel to
the plane passing through the others.

Let AB, BC,. two straight lines meeting one another, be parallel to
DE, EF that meet one another, but are not in the same plane with
AB, BC : The planes through AB, BC, and DE, EF shall not meet»
though produced. ' -

From the point B draw BG perpendicular (10. 2. Sup.) to the plane
which passes through DE, EF, and let it meet that plane in G ; and
through G draw GH parallel to ED (31. 1.), and GK parallel to EF :

188

SUPPLEMENT TO THE ELEMENTS

And because BG is perpendicular to the plane through DE, EF> it
must make right angles with
every straight line meeting
It hi that plane (L def. 2.
Sup.). But the straight lines S
QHy GK in that plane meet
it : Therefore each of (he
angles BGH, BGK is a right
angle : And because BA is
parallel (8. 2. Sup.) to GH
(for each of them is paral-
lel to DE), the angles GBA,
BGH are together equal

(29. I.) to two right angles : And BGH is a right angle ; therefore
also GBA is a rightangle, and GB perpendicular to B A : For the saiae
reason, GB is perpendicular to BC : Since, therefore, the straight
lin^ GB stands at right angles to the two straight lines BA, BC, that
cut one another in B ; GB is perpendicular (4. 2. Sup.) to the plane
through BA, BC : And it is perpendicular to the plane through DE»
EF ; therefore BG is perpendicular to each of the planes through
AB, BC, and DE, EF : But planes to which the same straight line
is perpendicular, are parallel (12. 2. Sup.) to one another : There-
fcre the plane through AB, BC, is parallel to the plane through DE>
EF. Wherefore, if two straight lines, &c. Q. E. D.

* Cob. It follows from this demonstration, that if a straight line meet
two parallel planes, and be perpendicular to one of them, it must be
perpendicular to the other also.

PROP. XIV. THEOR.

^iwo parallel plemes be eui hy another plancy thevr common teetiorur

vnth it are paraUels.

Let the parallel planes AB, CD, be cut by the plane EFHG, and
let their common sections with
it be EF, GH ; EF is parallel
foGH.

For the straight lines EF
and GH are in the same
plane, tiz. EFHQ which cuts
the planes AB and CD ; and
they do not meet though pro-
duced ;. for the planes in which
they arc do not meet ; there-
iore EF and GH are parallel
(def. 80. L). Q. E.D-

OF GEOMETRY. BOOK H.

189

PROP. XV. THEOR.

If two parallel planes he cut by a third plane, they have the same uidu

nation to that plane.

Let AB and CD be two parallel planes, and EH a third plane cut-
ting them ; The planes AB and CD are equally inclined to EH.

Let the straight lines EF and GH be the common section of the
.plane EH with the two planes AB and CD ; and from K, any point in
£F, draw in the plane EH the straight line KM at right angles to EF,
and let it meet GH in L ; draw also KN at right angles to EF in the
plane AB : and through the straight lines KM, KN, let a plane be
made to pass, cutting the plane CD in the line LO. And because EF
and GH are the common sections df the plane EH with the two pa-
rallel planes AB and CD, EF is parallel to GH (14. 2. Sup.)' But
EF is at right angles to the plane that passes through KN and KM
(4. 2. Sup.), because it is at right angles to the lines KM and KN :

therefore GH is also at right angles to the same plane (7. 2. Sup.),
and it is therefore at right angles to the lines LM, LO which it meets
in that plane. Therefore, since LM and LO are at right angles to
LG, the common section of the two planes CD and EH, the angle
OLM is the inclinalipn of the plane CD to the plane EH (4. def 2.
Sup.). * For the same reason the angle MKN is the inclination of the
plane AB to the plane EH. But because KN and LO are parallel,
being the common sections of the parallel pUmes AB and Cu with a
third plane, the interior angle NKM is equal to the exterior angle
OLM (29. 1.) ; that is, the inclination of the plane AB to the plane
EH, is equal to the indination of the plane uD to the same plane
EH. Therefore, fcc. Q. E« D.

lOO

SUPPLEMENT TO THE ELEMENTS

PROP. XVL THEOR.

If two strcdghi lines he cut hy parallel planes ^ they must he ad in the

same rcUio.

Let the straight lines AB, CD be cut by the parallel planes GH,
KL, MN, in the points A, E, B ; C, F, D : As AE is to EB, so is
CF to FD.

Join AC, BD, AD, and let AD meet the plane KL in the point X;
and join EX, XF : Because the two
parallel planes KL, MN are cut by the
plane EBDX, the common sections
EX, BD, are parallel (14. 2. Sup.).
For the same reason, because the
two parallel planes GB, KL are cut
by the plane AXFC, the common sec-
tions AC, XF are parallel : And be-
cause EX is parallel to BD, a side
of the triangle ABD, as AE to EB,
so is (2. 6;) AX to XD. Again, be-
cause XF is parallel AC, a side of the
triangle ADC,'as AX toXD, so is CF
to FD : and il was proved that AX
is to XD, as AE to EB : Therefore
(11. 6.), as AE to EB, so is CFto FD.
Wherefore, if two straight lines, &c.
Q. E. D.

PROP. XVIL THEOR.

If a straight line he at right angles to a plane, every plane which passes
through Ahai line is at right angles to the first mentioned plane, -

Let the straight line AB be at right angles to the plane CK ; every
plane which passes through AB is at right angles to the plane CK.

Let any plane DE pass through AB,and let CE be the common sec*
tion of the planes DE,CK ; take any point Fin CE, from which draw
FG in the plane DE at right angles to CE : And because AB is perpen*

dicular to the plane CK, therefore
it is also perpendicular to every
straight line meeting it in that plane
(1. def. 2. Sup.) ; and conse-
quently it is perpendicular to CE :
Wherefore ABF is a right angle ;
But GFB is likewise a right angle;
therefore AB is parallel (28.^ L)
to FG, And AB is at right angles
to the plane CK : therefore FG is

a A H

F B E

OF GEOMETRY. BOOK n.

101

also at right angles to the same plane (7. 2. Sup.)* Bat one plane is
at right angles to another plane when the straight lines drawn in one
of the planes, at right angles to their common section, are also at
right angles to the other plane (def. 2. 2.) ; and any straight line FG
in the plane D£, which is at right angles to CE, the common sectioii
of the planes, has been proved to be perpendicular to the other plane
OK ; therefore the plane DE is at right angles to the plane OK. In
like manner, it may be proved that all the planes which pass through
AB are at right angles to the plane CR. Therefore, if a straight line,
&c. a. E. D.

PROP- XVIII. THEOR.

if two planes cutting one another he each of them perpendicular to a third
ploTie, their common section is perpendicular to th6 same^lane.

Let the two planes AB, BO be each of them perpendicular to a
third plane, and BD be the common section of the first two ; BD is
perpendicular to the plane ADO.

From D in the plane ADC, draw DE perpendicular to AD, and DF
to DC. Because DE is perpendicular to AD, the common section of
the planes AB and ADC ; and because the ^

plane AB is at right angles to ADC, DE is . ^

at right angles to the plane AB (def. 2. 2.
Sup.), and thereA>re also to the straight line
BD in that plane (def. 1. 2. Sup.). For the
same reason, DF is at right angles to DB.
Since BD is therefore at right angles to both
the lines DE and DF, it is at right angles to
the plane in which DE and DF are, that is,
to the plane ADC (4. 2. Sup.). Wherefore,
&c. a. E. D.

A F

PROP. XIX. THEOR.

Tiro straight lines not in the same plane being given in position^ to draw
a straight line perpendicular to them hdh.

Let AB and CD be the given lines, which are not in the same
plane ; it is required to draw, a straight line which shall be perpea-

dicular both to AB and CD. „ , .^,^

In AB take any point E, and through E draw EF parallel to CD,
and let EG be drawn perpendicular to the plane which passes through
EB, BP (10. 2. Sup.). Through AB and EG let a plane pass, viz.

193 SUPPLEMENT TO THE ELEMENTS

GK, and let this plane meet CD in H ; from H draw HK perpendi^
calar to AB ; and HK is the line required. Through H, draw HO
parallel to AB.

^hen, since HK and GE, which are in the same plane, are hoth
at right angles to th^ straight line AB, they are parallel to one an-
other. And because the lines HG, HD are parallel to the lines EB,
£F, each to each, the plane GHD is parallel to the planes (13. 2.
Sup.) BEF ; and therefore EG, which is perpendicular to the plane
BEFy is perpendicular also to the plane (Cor. 13. 2. Sup.) GHD.
Therefore HK, which is parallel to GE, is also perpendicular to the
plane GHD (7. 2. Sup.), and it is therefore perpendicular to HD
(def. 1. 2. Sup.), which is in that plane, and it is also perpendicular
to AB ; therefore HK is drawn perpendicular to the two given lines,
AB and CD. Which was to be done.

PROP. XX. THEOR.

If a solid angle he contained by three plane angles, any two of these

angles are greater than the third.

Let the solid angle at A be contained by the three plane angles
BAC, CAD, DAB. Any two of them are greater than the third.

If the angles BAC, CAD, DAB be all equal, it is evident that any
two of them are greater than the third. But if they are not, let BAG
be that angle which is not less than either of the other two, and is
greater than one of them, DAB ; and at the
point A in the straight line AB, make in the
plane which passes through BA, AC, the
angle BAE equal (23. 1.) to the angle DAB ;
and make AE equal to AD, and through
E draw BEG cutting AB, AC in ^he points
B, Q, and join DB, DC. And because DA

is equal to AE, and AB is common to the -^^ ^

two triangles ABD, ABE, and also the an-

gle DAB equal to the angle EAB ; therefore the base DB is equai

Of* GEOMETRY. BOOK H. Ida

(4. 1.) to the base BE. And because BD, DC are greater (20. 1.)
than CB, and one of them BD has been proved equal to BE, a part
of OB, therefore the other DC is greater than the remaining part
EC. And because DA is equal to AE, and AC common, but the base
DC greater than the base EC ; therefore the angle DAC is greater
(25. 1.) than the angle EAC ; and, by the construction, the angle
DAB is equal to the angle BAE ; wherefore the angles DAB, DAO
are together greater than BAE, EAC, that is, than (he angle BAC«
But BAC is not less than either of the angles DAB, DAC ; there^
fore BAC;^ with either of them, is gfeater than the other. Where"
fore, if a solid angle, &c^ Q< E. D.

PROP. XXI. THEOR.

T%e plane angles which contain any solid angle ate together less ttutti

four right angles.

Let A be a solid angle contained by any number of plane aligles
BAC, CAD, DA£, EAF, FAB ; these together are less than four
right angles.

Let the planes which contain the solid angle at A be cut by another
plane, and letthe section of them by that plane be the rectilineal figure
BCDEF. And because the solid angle at B is contained by three plane
angles CBA, ABF, FBC, of which any
two are greater (20. 2. Sup.) than the
third, the angles CBA, ABF are great^.
er than the angle FBC s For the same
reason, the two plane angles at each of
the points C, D, E, F, viz. the angles
which are at the bases' of the triangles
having the common vertex A, are great-
er than the third angle at the same point,
which is one of the angles of the figure
BCDEF : therefore all the angles at the
bases of the triangles are together great-
er than all the angles of the figure : and
because all the angles of the triangles are together equal to twice as
many right angles as there are triangles (32. I.) ; that is, as there are
sides in the figure BCDEF ; and because all the angles of the Hgnre^
together with four right angles, are likewise equal to twice as many,
right angles as there ai'e sides in the figure (cor. L 32. L) ; therefore
all the angles of the triangles are equal to all the angles of the rectili-
tieal figure, together with four right angles. But all the angles at the
bases of the triangles are greater than all the angles of the rectilineal,
as has been pfoved. Wherefore, the remaining angles of the trian-
gles, viz. those at the vertex, which contain the solid angle at A, are
Jc88 than four right angles. Therefore every solid angle, &c. Q. £• D.

IM SUPPLEMENT TO THE ELEMENTS, ^bc.

Otherwise*^

Let the sum of all the angles at the bases of the triangles » S ^
die sura of alt the angles of the rectilineal 6gure BCD£Fs=2 ; the*
sum of the plane angles at A~X, and let R =s a right angle*

Then, because S+X^twice (32. 1.} aa many right angles as there-
are triangles, or as there ar^ sides of the rectilineal figure BCDEF^
and as 24-4R is also equal to twice as many right angles as there-
are sides of the same figure ; therefore S+X=r2+4R. fiut be«
cause of the three plane angles which contain a solid angle, any two-
are greater than the third, S72 ; and therefore XZ.4R ; that is, the-
sum of the plane angles which contain the solid angle at A is iesa-
than four right singles^ J^ £• D*

SCHOLIUM.

It is evident, that when any of the angles of the figure BCDEF i»
exterior, like the angle at D, in .
the annexed figure, the reasoning
in the above proposition does not
hold, because the solid angles at
the base are not all contained by
plane angles, of which two belong
to the triangular planes, having
their common vertex in A, and the
third is an interior angle of the
rectilineal figure, or base. There-
fore, it cannot be concluded that
S is necessarily greater than 2. Thia propositioD, therefore, is sub-
ject to a limitation, which is farther explained- ia the notea oa- this*
book»

ELEMENTS

GEOMETRY.

SUPPLEMENT.

BOOK III.

OF THE COMPARISON OF SOLIDS.

DEFINITIONS,

A Solid is that -which has length, breadth, and thickness.

'Similar solid figures are such as are contained by the same numl>er
of similar planes similarly situated, and having like inclinations to
one another.

IIL

A p3nramid is a solid figure contained by planes that are constituted
betwixt one j>lane and a point above it in which they meet.

IV-

A prism is a solid figure contained by plane figures, of which two that
are opposite are equal, similar, and parallel to one another ; and
the others are parallelograms.

A parallelopiped is a solid figure contained 'by six quadrilaterid
figures, whereof every opposite two are parallel.

A cube 4s a solid £gure contained -by six equal squares.

106 SUPPLEMENT TO THE ELEMENTS

VIL

A sphere is a solid figure described by the rerolution of a semicircle
about a diameter, which remains unmoved.

VUI.

The axis of a sphere is the fixed straight line about which the semi^
circle revolves^

IX.

The centre of a sphere is the same with that of the seniicircle.

The diameter of a sphere is any straight line which passes through
the centre, and is terminated both ways by the superficies of the
-sphere.

A cone is a solid figure described by the revolution of a right angled
triangle about one of the sides containing the right angle, which
side remains fixed.

XIL

The axis of a cone i^ the fixed straight line about which the triangle
revolves.

' . . XIII.

The base of a cone is the circle described by that side, containing
the right angle, which revolves.

XIV.

A cylinder is a solid figure described by the revolution of a right an-
gled parallelogram about one of its sides, which remains fixed.

XV.

The axis of a cylinder is the fixed straight line about which the pa«
rallelogram revolves.

XVL

The bases of .a cylinder are the circles described by the tjro revolv-
ing opposite sides of the parallelogram.

XVII.

Similar conos and cylinders are those which have their axes, and the
dmfnoiers of their buses proportionals.

OF GEOMETRY. BOOK IH.

lOT

PROP. I. THEOR-

If two solids he contaiiied hy the same number of equal and similar
planes similarly sittuU^d, and if the inclination of any two contiguous
planes in the one solid be the same with the inclination of the two
equal, and similarly situated planes in the other, the solids themselves
are equal and similar.

" Let AG and KQ be two solids contained by the same number of
equal and similar planes, similarly situated so that the plane AC is si.
niilar and equal to the plane KM, the plane AF to the plane KP ; BG
to LQ, GD to QN, DE to NO, and FIl to PR. Let also the incli-
nation of the plane AF to the plane AC be the same with that of the
plane KP to the plane KM, and so of the rest ; the solid KQ is equal
and similar to the solid AG.
, Let the solid KQ be applied to the solid AG, so that the bases KM

and AC, which are equal and similar, may coincide (8. Ax. 1.), the
point N coinciding with the point D, K with A, L with B, and so on.
And because the plane KM coincides with the plane AC, and, by hy-
pothesis, the inclination of KR to KM is the same with the incKna-
lion of AH to AC, the plane KR will be upon the plane AH, and will
coincide with it, because they are similar and equal (8. Ax. L), and
because their equal sides KN and AD coincide. And in the same
manner it is shewn that the other planes of the solid KQ coincide
with the other planes of the solid AG, each with each : wherefore
the solids KQ and AG do wholly coincide, and are equal and similar
to one another. Therefore, &c. Q. £. D.

PROP. IL THEOR.

Jf a solid be contained by six planes, two and .two of which are parallel^
tJui opposite planes are similar and equal parallelograms.

Let the solid CDGH be contained by the parallel planes AC, GF ;
BG, CE ; FB, AE : its opposite plahes are similar and equal paral-
klograms.

Because the two parallel planes BG, CF, are cut by the plane AC,

1«8

SUPPLEMENT TO THE ELEMENTS

^their common sections AB, CD are pafallel (14. 2. Sup.). Again,
because the two parallel planes BF, AE are cut by the plane AC,
their common sections AD, HC are parallel (14. 2. Sup.) : and AB is
parallel to CD ; therefore AC is a parallelogram. In like manner, k
may be proved that pach of the figures £ ji

CE, FG, GB, BF, AE is a parallelogram ; ^

Join AH, DF ; and because AB is parallel J^
to DC, and BH to CF ; the two straight
lines AB, BH, which meet one tinother,
are parallel to DC and CF, which meet
one another ; wherefore, though the first
two are not in the same plane with the
other two, they contain equal angles (9« 2. TJ
Sup.) ; the angle ABH is therefore equal to the angle DCF. And
because AB, BH, are equal to DC, CF, and the angle ABH 6qual to
the angle DCF; therefore the base AH is equal (4. 1.) to the base
DF, and the triangle ABH to the triangle DCF: For the same rea-
son, the triangle AGH is equal to the triangle DEF : and therefore
the parallelogram BG is equal and similar to the "parallelogram CE.
In the same manner, it may be proved, that the parallelogram AC is
«qual and similar to the parallelogram GF, and the parallelogram AE
to BF. Therefore, if a. solid, <&;c. Q. E. D.

PROP. IIL THEOR.

If a solid parallelopiped be cut by a platie parMd to ttoo of its opposite
planes, it loill be divided into two solids^ which will be to one another
-as the bases*

Let the solid parallelopiped ABCD be cut by the plane EV, which
is parallel to the opposite planes AR, HD, and divides the whole into
the solids ABFV, EGCD ; as the base AEFY to the base £UCF,80
is the solid A BFV to the solid EGCD.

Produce AH both ways, and take any number of straight lines HM»
MN, each equal to EH, and any number AK, KL each equal to EAp
flud complete the parallelograms LO, KY^ HQ, MS, and the solids LP#

Or 1

Ny|Np|Xyr\ T

^^^

a s

Y r c

KR, HU, MT ; then, because the straight lines LE, EA, AE are all

OP GEOMETRY. BOOK HI. lOff

equal, and also the straight lines KO, AY, £F which make equal an-r
gles with LK, KA, AKfSie parallelograms LO, KY, AF are equal and
similar (36. 1. 6c def. 1. 6.) : and likewise the parallelograms KXy
KB, AG ; as also (2. 3. Sup.) the parallelograms LZ, KP, AR, be*
cause they are opposite planes. For the same reasos, the parallel*
ograms EG, HQ, MS are equal (36^ 1. 6c def. 1. 6.) ; and the paraW
lelograms HG, HI, IN, aJi also (2,3. Sup.) HD, MU, NT ; therefore
three planes of the solid LP, are equal and similar to three planes
of the solid KR, as also to three planes of the solid A V : but the three
planes opposite to these three are equal and'similar tothem (2. 3. Sup )
in the several solids f therefore the solrds LP, KR, AV are contained
by equal and similar planes. And because the planes LZ, KP, AR
are parallel, and are cut by the plane XV, the inclination of LZ to XP
is equal to that of KP to PB.; or of AR to BY (15. 2. Sup.) and the
same is true of the other contiguous plafies, fherefore the solids LP,
EIR, and AV, are equal to one another (1. 3. Sup.). For the same
reason, the three solids ED, HU, MT are equal to one another ; there-
fore what multiple soever the base LF is of the base AF, the same
multiple is the solid LV oTthe solid AV ; for the same reason, what-
ever multiple the base NF is of the base HF, the same multiple l»
the solid NV of the solid ED : And if the base LF be equal to the
base NF, the solid LV is equal (1. 3. Sup.) to the solid NV-; and if
the base LF be grediter than the base NF, the solid LV is greater thaa
the solid NV\: and if less, less. Since then there are four magnitudes,,
viz. the two bases AF, FH, and the two solids AV, ED, and of the
base AF and solid AV, the base LF and solid LV are any equimul-
tiples whatever; and of the base FH and solid £D, the base FN and
solid NV are any equimultiples whatever ; and it has been proved,
that if the base LF is greater than the base FN, the solid LV is*
greater than the solid NV ; and if equal, equal ; and if less, less i
Therefore (def. 5. 5.) as the base AF is to the base FK, so is the
solid AV to the solid ED. Wherefore, if a solid, dz;c. Q. E. D.

Cos. Because the parallelogram AF is to the parallelogram FH
as YF to FC (L 6.)« therefore the solid AV is to the solid ED as YF
to FC.

PROP. IV. THEOR.

j^ a solid p€ur€^lopiped be cut hy a plane passing tKrougn the dhagonalsf
of two of the opposite planes, it will be cut into two equcd prisms.

Let AB be a solid parallelepiped, and DE, CF the diagonals of the
opposite parallelograms AH, GB, viz. those which are drawn betwixt
the equal angles in each ; and because CD, FE are each of them pa-
vallel to GA, though not in the same plane with it, CD, F£ are pa*
fallel (^. 2. Sup.) ; wherefore the diagonals CF, DE are in the plane

300 SUPPLEMENT TO THE ELEMENTS

in which the parallels are, and are tbem-
selves pal-albls (14. 2. Sup.) ; ihe plane
CDEF cula the solid AB inio two equal parts.

Because the triaojcle CGF ia equal (34.
I.) to the triangle CBF, and the triangle 0-^
DAE to DHE ; and since the parallelogram
CA ia equal (2. 3. Sup.) and similar to the
opposite one BE ; and the parallelogram GE
lo CH : therefore ihe planes which contain
the prisms CAE, CBE, are equal and srmi.

lar, each to each ; and they are also equally

inclined lo one another, because the planes A.
AC, EB are parallel, as also AF and BD, and they are cut by the
plane CE (15. 2. Sup.). Therefore the prism CAE is equal to the
prisRi-CBE (1. 3. Sup.), and the solid AB is cut into two et|ual
prisms by the plane CDEF. Q. E. D.

N. B. The insisting straight tines of a parallelopiped, mentioDed
, in the fallowing propositions, are the sides of the parallelograms bv
twist the base and Ihe plane paraliel to it.

PROP. V. THEOR.

Solid paraUelopipeda ttpon the same base, and of the lame ahitude, thi
insisling straight lines of which are terminaled in tie same straight
lines in the plane opposite ft? the base are equal to one another.

be upon the same base AB,
listing straight lines AF, AG,
light line FN, and let Ihe in--
iJnated in the same straight
nlid AK.

QB is«qual (34. 1.) to each
reDHiseqii^lloEK: add,
DEisequajtoHK: Where-
..) to the-l^iJingle EHK: and
the paraHe.ttjgram HN, For
al to the. triangle. LMN, an^
the parallelogram CF is equal'(2. 3. Sup.) to the parallelc^ram BM,
and GG to BiV ; for they are opposite. Therefore ifae planes whicb

OP GEOUETHT. BOOK ID. SOt

ctMataiii tke prism DAG are similur and equal to thoise which contain
f be prism HLN, each to each ; and the contignous planes are also
equally inclined to one another (15. 2. Sup.), because that the paraK
!ef planes AD and LH, as also ASS and LK, are cut by the same plane
DN : therefore the prisms DAG, HLN are equal (1. 3. Sup*)% If
therefore the prism LNH be taken from the solid, of which the base
is the parallelo^am AB, and FDK!N the plane opposite to th^ hza^ ^
and if rrora this same solid there be taken the prism AGD, the remain-
ing solid, viz. the parallelepiped AH is equal to the remaining pand^
lelopiped AK. Therefore solid paralleloprpeds, d;c. Q. £. D.

PROP, VI, THEOR.

Solid paraUelopipeds tiponthe same base, and ^fthesamedUUudey the tn-
wUng straight lines of which are not terminated in the same sltaighit
lines in the ]^ane opposite to the base, are equal to one another^

Xet the parallelopipeds C]VI,/CN', be upon the same base AB» amd
of the same altitude, but their insisting straight, lines AF, AG, LM,
LiN, CD, CE, BH, BK, not terminated in t^ same straight lines ;
the solids CM, CN are equ^l to one another.

Produce FD, MH, and NG, KE>and let them meet one another in
the points 0, P, Q, R ; and join AO, LP, BQ, CR. Because the pianos
{4^ 5. 3. Sup.), LBHM and ACDF are parallel, and because the

)>kne LBHM is that in which are the parallels LB, MHPQ (def. 5. 3»
Bup.), and in which also is the figure BLPQ ; and because the plane
ACDF is that in' which a^e the parallels AC, FDOR, and in which also
is the figures C AOR ; therefore the figures BLPQ, CAOR, are in pa-
rallel planes. In like manner because the planes ALNG and CBKB
are parallel, and the plane ALNG is that in which are the parallels Ah.
OPGN, and in which also is the figure ALPO ; and the plane CBKB
is that in which are the parallels CB, RQEK, eind in which also is th«

809

SUPPLEMENT TO THE ELEMENTS

figure GBQR ; therefore the figures ALPO, CBQR, are ia parallel
planes. But the planes ACBL, ORQP are also parallel ; therefore
the solid CP is a parallelopiped. Now the solid parallelepiped CM is
equal (5. 2. Sup.) to the solid parallelepiped CP, because they are
upon the same base, and their insisting straight lines AP, AO, CD,
CR ; LM, LP, BH, BQ are terminated in the same straight lines FR,
MQ ; and the solid CP is equal (5. 2. Sup.) to the solid CN ; for they
are upon the same base ACBL, and their insisting straight lines AO,
AG, LP, LN ; CR, C£, Bii, BK are terminated in the same straight
lines ON, RK ; Therefore the solid CM is equal to the solid CN.
Wherefore solid parallelepipeds, &c. Q. £. D.

PROP, VIL THEOR.

ScUd paraUdopipedsj which are upon equal basesy and of the same alii'

tude, are equal to one another*

Let the solid parallelepipeds, AE^ CF, be upon equal bases AB, CD,
and be of the same altitude ; the solid AE is equal to the solid CF*

Case L Let the insisting straight lines be at right angles to the bases
AB, CD, and let the bases be placed in the same plane, and so as that
the sides CL, LB> be in a straight line; therefore the straight line LM,
which is at right angles to the plane in which the bases are, in the
point L, is common (11. 2. Sup.) to the tWb solids AE, CF ; let the
other insisting lines of the solids be AG, FIK, BE ; DF, OP, CN : and
first, let the angle ALB be equal to the angle OLD ; then AL, LD are
in a straight line (14. 1.). Produce OD, HB, and let them meet in Q
and complete the solid parallelepiped LR,the base of which is the pa-
rallelogram LQ, and of which LM is one of its insisting straight lines :
therefore, because the parallelogram AB is equal to CD, as the base
AB is to the base LQ, so is (7. .5.) the base CD to the same LQ : and
because the solid parallelepiped AR is cut by the plane LMEB, which
ia parallel to the opposite planes AK, DR ; as the base AB is to the

A S

H T

base LQ, so is (3. 3. Sup.) the solid AE to the solid LR : for the same
reason because the solid parallelepiped CR is cut by the plane LMFD,
which is parallel to the opposite planes CP» BR ; as the base CD to

OF GEOMETRY. BOOK HL 908

Ihie base LQ ; so is the solid CF to the solid TjR ; but as the base AB
to the base LQ, so the base CD to the base LQ» as has been proved :
therefore as the solid AE to the solid LR, so is the solid CF to the solid
LR ; and therefore the solid A£ is equal (9. 5.) to the solid CF.

But let the solid paralleloptpeds, SE, CF be upon equal bases SB,
CD, and be of the same altitude, and let their insisting straight lines be
a^ right angles to the bases ; and place the bases SB, CD in the same
p4ane, so that CL, LB be in a straight line ; and let the angles SLB,
OLD be unequal ; the solid SE is also in this case equal to the solid
CFi Produce DL, TS until they meet in A, and from B draw BH
parallel to DA^ and let HB, OD produced meet in Q, and complete
the solids AE, LR : therefore the solid AE, of which the base is the
parallelogram LE, and AK the plaile opposite to it, is equal (5. 3. Sup.)
to the solid SE, of which the base is LE, and SX the plane opposite ;
for they are upon the same base LE, and of the same altitude, and
their insisting straight lines, viz. LA, LS, RH, BT ; MG, MU, EK,
EX, are in the same straight lines AT, GX : and because the parallel,
ogram AB is equal (35. 1.) to SB, for they are upon the same base
LB, and between the same parallels LB, AT ; and because the base
SB is equal to the base CD ; therefore the base AB is equal to the
base CD : but the angle A LB is equal to the angle OLD : therefore,
by the first case, the solid AE is equal to the solid CF ; but the solid
AE is equal to the solid SE, as was demonstrated : therefore the solid
SE is equal to the solid CF*

Case 2. If the insisting straight lines AG, HK, BE, LM ; CN, RS,

• T

DF, OP, be not at right angles to the bases AB, CD ; in this case
likewise the solid AE is equal to the solid CF. Because solid para.1-
lelopipeds on the same base, and of the same altitude, are equal
(6. 3. Sup.), if two solid parallelopipeds be constituted on the bases
AB and CD of the same altitude with the solids AE and CJF, and with
their insisting lines perpendicular to their bases, they will be equal
to the Solids AE and GF ; and, by the first case of this proposition,
they will be equal to one another? wherefore, the solids AE and CF
iure also equal. Wherefore, solid parallelepipeds, ^c. Q. E. D, ^

SUPPLEMENT TO THE ELEMENTS

PROP. VIIL THEOR.

Solid pamttdopipeds which have the same aititudey are fo tme another

ae their haaea.

Let AB, CD be sofid paralleloptpeds of the same altitude : they
9re to one another as their bases ; that is» aa the base AE to the b€tfe
CF, so is the solid AB to the solid CD.

To the straight fine FG apply the parallelogram FH equal (Cor.
45. 1.) to A£, so that the angle F6H be equal to the angle LCG ;

and eomplete the solid parallelepiped GK upon the base FH, one of
whose insisting lines is FD, whereby the solids CD, GK must be of
the same altit«ide» Therefore the solid AB is equal (7. ^ Sup.) to
the solid GK, because they afre upon equal bases AE, FH, and are of
the same altitude : and because the solid parallelepiped CK is cut by
the plane DG which is parallel to its opposite planes, the base HF is
(9. 3. Sup.) to the base FC, as the solid HD to the solid DC : But
the base HF is equal to the base AE, and the solid GK to the solid
AB : therefore, as the base AE to the base CF, so is the solid AB to
the solid CD. Wherefore solid paraUelopipeds^ dec. Q. E. D.

CoR. 1. From this it is manifest, that prisms upon triangular bases^
and of the same altitude, are to one another as their bases. Let the
prisms BNM, DPG, the bases of which are the triangles AEM, CFG,.
have the same altitude : complete the parallelograms AE, CF, and the
solid parallelepipeds AB, CD, in the first of which let AN, and in the
other l^t CP be quo of the insisting lines.^ And because the solid pa-
rallelopipeds AB, CD have the i^me altitude, they are to one another
as the base AE is to the base CF ; wherefore the prisms, which are
their halves (4i 3. Sup.) are to one another, as the base AE to the
base CF ; that is, as the triangle AEM to the triangle CF€L

Cor. 2. Also a prism and a parallelepiped, which have the same a?-
titude, are to one another as their bases ; that is, the prism BNM is
to the parallelopiped CD as the triangle AEM to the parallelogran
LG. For by the last Cor. the prism BNM is to the prism DPG as the
triangle AME to the triangle CGP, and therefore the prism BNM is t»
twice the prison DPG as die triangle AME to twice the triangle CGP

OF GEOMETRY. BOOK III.

205

(4. 5.) ; that is, the prism BNM is' to the parallelopiped CD as the
triangle AME to the parallelogram .LG.

PROP. IX. THEOR.

SdH pardUelopipeds are to one anther in theraiio that is compounded
of the ratios of the areas of their bases^ and of their altiiudes.

laCt AF and GO be two solid parallelepipeds, of which the bases
are the parallelograms AG and GK, and the altitudes, the perpen-
diculars let fall on the planes of these bases from any point in the
opposite planes £F and MO ; the solid AF is to the solid GO in a
ratio compounded of the ratios of the base AC to the baipie GK, and
of the perpendicular on AC, to the perpendicular on GK.

Case 1. When the insisting lines are perpendicular to the bases
AC and GK, or when the solids are upright.

In GM, one of the insisting lines of the solid GO, take GQ equal to
AE, one of the insisting lines of the solid AF, and through Q let a
plane pass parallel to the plane GK, meeting the other insisting lines

of the solid GO in the points B, S and T. It is evident that GS is a ^
solid parallelopiped (def. 5. 3. Sup.) and that it h^s the same altitude
with AF, viz. GQ or AE. Now the solid AF is to the solid GO in a
ratio compounded of the ratios of the solid AFtothe solid GS (def. 10.
5.), and of the solid GS to the solid GO ; but the ratio of the solid
AF to the solid GS, is the same with that of the base AC to the base
GK (8. 3. Sup.), because their altitudes AE and GQ are equal ; and
the ratio of the solid GS to the solid GO, is the same with that of GQ
to GM (3. 2. Sup.) ; therefore, the ratio which is compounded of the
ratios of the solid AF to the solid GS, and of the solid GS to the solid
GO, is the same with the ratio which is compounded of the ratios of
the base AC to the baseGK, and of the altitude AE to the altitude GM
(F.,5.). But the ratio of the solid AF to the solid GK), is that which
is compounded of the ratios of AF to GS, and of GS to GO ; therefore,

2d6 SUPPLfiMfiNT TO THE ELEMENTS

tlie ratio of the sqlid AFto the soh'cl GO is eompounded of theratiofei aX
the base AC to the base GK, and of the ahitude A£ to the altitude 6M»

Case 2. When the insisting lines are not perpendicular to the bases.

Let the parallelograms AC and GK be the bases as before, and let
AE and GM be the altitudes of two parallelepipeds Y and Z on these
bases. Then, if the upright parallelepipeds AFand GO be constitute
ed on the bases AC apd GK» with the altitudes AE and GM, they will
be equal to the parallelopipeds Y and Z (7. 3. Sup.). Now, the solids
AF and GO, by the f^rst case, are in the ratio compbuhded of the
ratios of the bases AC and GK, and of the altitudes AE and GM ;
therefore also the solids Y and Z have to one another a ratio that is
compounded of the same ratios. Therefore, &c. Q. E. D.

Cor. 1. Hence, two straight lines may be found having the same ra*
tio with the two parallelopipeds AF and GO. To AB, one of the sides
of the parallelogram AC, apply the parallelogram BV equal to GK,
having an angle equal to the angle BAD (44. 1.) ; and as AE to GM, so
let AV be to AX (12. 6.), then AD is to AX as the solid AF to the
solid GrOr For the ratio of AD to AX is compounded of the ratios
(def. 10. 6.) of AD to AV, and of AV to AX ; but the ratio of AD to
AV is the same with that of the parallelogram AC to the parallelo-
gram BV (1. 6.) or GK ; and the ratio of AV to AX is the same with
that of AE to GM ; therefore the ratio of AD to AX is compounded of
the ratios of AC to GK, and of AE to GM (E. 5.). But the ratio of
the solid AF to the solid GO is compounded of the same ratios ;
therefore, as AD to AX, so is the solid AF to the solid GO.

Cor. 2. If AFand GO are two parallelopipeds, and if to AB, to the
perpendicular from A upon DC, and to the altitude of the parallele-
piped AF, the numbers L, M, N be proportional : and if to AB, to
OH, to the perpendicular from G on LK, and to the altitude of the
parallelepiped GOi the numbers L, Z, m, n be proportional ; the solid
AF is to the solid GO as L XM XN to Z X«i Xn.

For it may be proved, as in the 7th of the 1st of the Sup. that L x
MXN is to ZXmXn in the ratio compounded of the ratio of LXM
to ZXm, and of the ratio of N to n. Now the ratio of LxM to lyim
is that of the area of the parallelogram AC to that of the parallelo.
gram GK ; and the ratio of N to n is the ratio of the altitudes of the
parallelopipeds, by hypothesis, thierefore, the ratio of LXMXN to
ZXmXtt is compounded of the ratio of the areas of the bases, and of
the ratio of the altitudes of the parallelopipeds AF and GO ; and the
ratio of the parallelopipeds themselves is shewn, in this proposition,
to be compounded of the same ratios ; therefore it is the same with
that of the product LxMxN to the product ZXwXn.

CoR. 3. Hence all prisms are to one another in the ratio' com-*
pounded of the ratios of their bases, and of their altitudes. For every
prism IS equal to a patallelopiped of the same altitude with it, and of
an equal base (2. Cor. 8. 3. Sup*},

OP GEOMETRY. BOOK HI.

PROP. X. THEOR.

Solid paraillelopipeds, which have their bases and aliiiudes reetprocatty
. proportional^ are equal ; and 'pardUelopipeds which are equals have
their hoses and altitudes reciprocally proportional. ^

Let AG and KQ be two solid parallelopipeds^ of which the bases
are AC and KM, and the altitudes AR and KO, and let AC be to
KM as KO to AE ; the solids AG and KQ are equal.

As the base AC to the base KM, so let the straight line KO be to
the straight line S. Then, since AC is to KM as KO to S, and also

3EL

]sr

A
S

— ^

by hypothesis, ^C to KM as KO to AE, KO has the same ratio to S
that it has to AE (11. 5.) ; wherefore AF is equal to S (9. 5.). But
the solid AG is to the solid KQ, in the ratio compounded of the ratios
of AE to KO, and of AC to KM (9. 3. Sup.) that is, in the ratio com-
pounded of the ratios of AE to KO, and of KO to S. And the ratio
of AE to S is also compounded of the same ratios (def 10. 6.) ; there-
fore, the solid AG has to the solid KQ the same ratio that AE has to
S. But AE was proved to be e4ual to S, therefore AG is equal to
KQ.

Again, if the solids AG and KQ be equal, the base AC is to the
base KM as the altitude KO to the altitude AE. Take S, so that AC
may be to KM as KO to S, and it will be shewn, as was done above,
that the solid AG is to th^ solid KQ as AE to S ; now, the solid AG
is, by hypothesis, equal to the solid KQ : therefore, AE is equal to
S ; but, by construction, AC is to KM, as KO is to S'; therefore,. AC
is to KM as KO to AE. Therefore, &c. Q. E. D,

Cor. In the same manner, it may be demonstrated, that equal
prisms have their bases and altitudes reciprocally proportional, and
conversely.

PROP- IX. THEOIL

Similar sdUd paraUdopipeds are to one another in the triplicate ratio

of their homologous sides.
Let AG, KQ be two similar parallelepipeds, of which AB uul KL
are two homologous sides ; the ratio of the solid AG to the sond KQ
is triplicate of the ratio of AB to KL.

1108

SUPPLEMENT TO THE ELEMENTS

Because the solids are similar, the parallelograms AF, KB are simi-
lar (def. 3. 3. Sup.), as also the parall^lognuns AH, KR ; therefore,

> \

IPv

L .

the ratios of AB to KL, of AE to KO, and of AD to iCN are all equal
(def. 1. 6.). But the ratio of the solid AG to the solid KQ is conii
pounded of the ratios of AC to KM, and of AE to KO. Now, the
ratio of AC to KM, because they are equiangular paraJlelograois, is
compounded (23. 6.) of the ratios of AB to KL, and of AD to RN.
Wherefore, the ratio of AG to KQ is compounded of the three ratios
of AB to KL, AD to KN, and AE to KO : and the three ratios have
already been proved to be equal ; therefore, the ratio that is com*
pounded of them, viz. the ratio of the solid AG to the solid KQ, is
triplicate of any of them (def. 12. 5.) : it is therefore triplicate of the
ratio of AB to KL. Therefore, similar solid parallelopipeds, d^c.
Q. E. D. '

•■ .

Cob. 1. If as AB to KL, so KL to m, and as KL to m, so is m to n,
then AB is to n as the solid AG to the solid KQ. For the ratio of A]3
to n is triplicate of the ratio of AB toKL (def 12.5.), and is therefore
equal to that of the solid AG to the solid KQ.

Cor. 2. As cubes are similar solids^ therefore the cube on AB is to
the cube on KL in the triplicate ratio of AB to KL, that is in the same
ratio with the solid AG, to the solid KQ. Similar solid parallelopipeds
are therefore to one another as the cubes on their homologous sides.

Cor. 3. In the same manner it is proved, that similar prisms are to
one another in the triplicate ratio, or in\he ratio of the cubes of their
homologous sides.

PROP. XIL THEOR.

If two triangular pyraniidsy which have eqwd bases and altitudes^ he
cut by pla7ies that are parallel to the bases ^ and at equcd distances
^ fnym them, the sections are equal to one another*

Let ABCD and EFGH be two pyramids, having equal bases BDG
and FGH| and equal altitudes, vis* the perpendiculars AQ, and Ed

^t GEOMETRYw BOOK IIL

C F

^a^a from A and E upoa the pkines BDC and FGH : and let them
he cut by planes parallel to 6D0 and FGH, and at equal altitudes
QR and ST above those planes, and let the sections be the triangles
ICLM, NOP ; KLM and NOP are equal to one another.

Because the plane ABD cuts the parallel planes BDG, KLM^ the
cctomon sections BD and KM are paraUel (14. 2. Sup.}. For the
same reason, DC and ML are parallel. Since therefore KM and
ML are parallel to BD and DC, each to each, though not in the same
plane with them, the angles KML is equal to the angle BiyG {9. 2^
Sup.). In like manner the other. angles of thesfe triangles are prov-
ed to be equal ; therefore, the triangles are equiangular, and cons6<i
^quently similar ; and the same is true of the triangles NOP, FGH«.

Now, since the straight lines ARQ, 4RB meet the parallel planes
6DC and KML, they are cut b^ them proportionally (16. 2. Sup.%
or QR : RA : : BK : KA ; and AQ : AR : : AB : AK (1& 5.), fot
the same reason, £S : ET : : £F : £N ; therefpre AB : AK : i
£F : EN, because A^ is equal to ES, and AR to ET. Again* be^
<cause the triangles ABC, AKL arc similai^

AB ^ AK : : BC : KL ; and for the same reasott
EF : EN : : FG : NO ; therefore,

IBC : KL : : FG : NO. Attd» when ibur straight lines are prd^
|>ortionals, the similar figures described on them are also proportion^
als (22. 6. j ; therefore ^e triangle BCD is to the triangle KLM as
ihe triangle FGH to the triangle NOP ; but the triangles BDC, FGH
are equd ; therefore, the triangle KLM is also equal to the triaikigle
NOP (1. 6.). Therefore, &c. GL E. D.

Go&. 1. Because it has been jshewn that the triangle KLM is si-
milar to the base BCD ; therefore, any section of a triangular py.
ramid parallel to the base, is a triangle similar to the base. And in
ihe same manner it is shewn, that the sections parallel to the base
of a polygonal pyramid are similar to the base.

CoE. 2. Hence also, in polygonal pyramids of equal bases and a]a«

SIO SUPPLEMENT TO THE EI^MENTS

tadea, the nctioiu parallel to the batei, and at eqtiat distances front
Ifaem, are equal to one another.

PROP. XIII. THEOR.

Aterieaofiprumi of the tame abitude may 6e circMiuerihed abmU my
pyramid, stLch that the turn of the priamt »haU exceed the pyramid bjf
a idid las than any giixn tolid.

Let ^BCD be a pyramid, and Z* a given aolid ; aseriesof pHanw
bavinff all the same altitude, may be circumscribed about the pyramid
ABCD, so that their sum shall exceed ABCD, by a aolid leas than Z.

Let Z be equal to a prism standing on the same base with the py-
ramid, viz. the triangle BCD, and .^y- -p,
having for its altitude the perpendi'
cular dra»D from a certain point E in
the line AC upon the plane BCD. It
ia evident, that CE multiplied by a
certain namber m will he greater than
AC ; divide CA into as many equal
parts as there are units in m, and let
these be OF, f G, GH, HA, each ef
whichwill be less than CE. Through
each of the points F, O, H, let planes
be made to pass parallel to the plane
BCD, making with the sides of the
pyramid the sections FPQ, GRS,
HTU, which will be all similar to one
another, and to the base BCD ( 1 ■ cor.
12. 3. Sup.). From the point B
draw in the plane of the triangle ABC,
the straight line BK parallel to CF
meeting FP produced in K. In like
manner, from D draw DL parallel to

CF, meeting PQ in L: Join KL, and it is plain, that the solid KBCDLF
Is a prism (def. 4. 3. Sup.). By the same constructton, let the prisms
PM, RO, TV be described. Also, let the straight line IP, which is in
the plane of the triangle ABC, be produced till it meet BC in h ; aod
let the line MQ be produced till it meet DC in g ; Join hg ; then
hC gQFP is a prism, and is equal to the prism PM ( 1 . Cor. B. 3. Sup.).
In the same manner is described the prism MS equal to the prism RO,
sad the prism qU equal to the prism TV. The sum, therefore, of all
Ihe inscribed prismshQ.mS, and qUia equal to the sum of the prisma
FM, RO and TV, that is, to the sum of all the circumscribed prisms
except the prism OL ; wherefore, BIr is the excess of the prism cir-
cumscribed about the pyramid ABCD above the prisms inscribed with

■ TtM Solid Z 1( not npreiented id the ^ae of ihit, or (b« ftBowieg FuipoaitlOa.

QF GEOMETRT. BOOK III.

in it. But the priam BL U leat than the prif m which hM the tritng !•
BCD for its haae^ and for its altitude the perpendicular from E upoa
ihe plane BCD ; and the prism which has BCD for its base* and the
perpendicular from E for its altitude^ is by hypothesis equal to the
given solid Z ; therefore the excess of the circumscribed, above the
inscribed prisms, is less than the given solid Z« But the excess of the
circumscribed prisms above the inscribed is greater than their ex-
cess above the pyramid ABCD, because ABCD is greater than the
sum of the inscnbed prisms. Much more, thereforCf is the excess
of the circumscribed prisms above the pyramid, less than the solid Z«
A series of prisms of the same altitude has therefore been ciretmiu
scgibed about the pyramid ABCD, exceeding it by a solid leii tbao
the given solid Z. Q. £• D.

PROP. XIV. THEOR.

Pyrmmids Ito have eqtud bates mnd 4dlUitd€s an equal to one aneiher*

Let ABCD, EFGH, be two pyramids that have equal bases BCD,
FGH, and also equal altitudes, viz* the perpeodieulars drawn froni
the Teitices A and £ upon the planes BCD, FGH s the pyfamid
ABCD is equal to the pyramid EFGH.

If tkey are not equal, let the pyratnid EP6H exceed the pyramid
ABCD by the solid 2L Tben,aeeriesofprisfniof(lieiaiiie aMode

,rT,

pyMKid AIK^D tlMt sbaM exeeed k, by «

than Z (13. S. ihip.; ; let HMsee be tW^MWM tfaatt bair« Ay

ts»a^e« IK^. l»i<^ OfU, I^H. iJ^^i^ KM i«^

nf Mwd Mrts tm^ which Ai> M^ivMled, via. ||T,

^^i ihe MMls Y, t Md ¥, let Hie eeci^

313

SUPPLEMENT TO THE ELEMENTS

TZW, UHX, y<l»Y be made parallel to the base FGH. The seetioif
KQL 18 equal to the section WZT (12. 9. Sup.); a^also ORI toXHU,
and FSM to Y4»V ; and therefore also the prisms that stand upon the
equal sections are equal (1. Cor. 8. 3. Sup.)^ that is, the prisnv which
stands on the base BCD, and which is between the planes BCD and
NQrL, is equal to the prism which stands on the base F6H, and which
is between the planes FGH and WZT ; and so of the rest, because
they have the same altilude: wherefore, the sum of all the prisme
described about the pyranud ABCD is equal to the sum of all those
described about the pyramid EFGH. But the excess of the prism&r
described about the pyramid ABCD abore the pyramid ABGD is less
Ihan Z (13. 3. Sup.) ; and therefore, the excess of the prism describ-
ed about the pyramid EFGH above the pyramid ABCD is also less
than Z. But the excess of the pyramid EFGH above the pyramid
ABCD is equal to Z, by hypothesis, therefore, the pyramid EFGH
exceed the pyramid ABCD, more than the prisms described about
EFGH^xeeed the same pyramid ABCD. The pyramid EFGH is
therefore greater than the sum of the prisms described about it»
which is impossible. The pyramids ABCD, EFGH, therefore, are
not unequal, that is, they are equal" to one another. Therefore,^
pyramids, &c. Q. E. D.

PRDP. XV. THEOR.

Every prism having a triangular hctse may he divided into iHree pyra-
mids thai haeoe triangidur bases, and that are equal to another.

Let there be a prism of which the base is the triangle ABC, and
let DEF be the triangle opposite the base : The prism ABCDEP.
may be divided into three equal pyramids- having triangular bases.

Join AE, EC, CD ; and because ABED is a parallelogram, of whieb
AE is the diameter, the triangle ADE is equal (34. 1.) to the triangle
ABEl I. therefore the pyramid of which the
base is the triangle ADE, and vertex the
point C, is equal (14. 3. Sup.) to the pyra-
Slid, of which the base is the triangle ABE,
and vertex the point C. But the pyramid j>
of which the base is the triangle ABE, and
vertex the point C, that is, the pyramid ABCE
»equal tethe pyramid DEFC (14. 3. Sup.),
Ibi tbey have equal bases, viz. the triangles^
ABC, DEF, and the same altitude, viz. the
^titude of the prism ABCDEF. Therefore
the three pyramids ADEC, ABEC, DFEC
are equal to one another. But the pyra-
iBiids ADEC, ABEC, DFEC make up the
whole prism ABCDEF ; therefore, the prism
^BCDEF is divided into three equal pyra. A
wtov WbiEirefore, &<:• Q, £• J>^

OP GEOMETRY. BOOK. HI.

218

_ _ •

Cos. 1. From this it is manifest, that every pyramid is the third
part of a prism which has the same base, and the same altitude with
it ; ibr if the base of the prism be any other figure than a triangle,
it may be divided into prisms having triangular bases.

CoR. 2. Pyramids of equal altitudes are to one another as their
bases ; because the prisms upon the same bases, and of the isaine
altitude, are (1. Cor. 38. Sup.) to one another as their bases.

PROP. XVI. THEOR-

If from any point in the circumference of the base of a cyHndety a
straight line he drawn perpendicular to the plane of the hase^-it wiU
be wholly in the cylindric superficies.

Let ABCD be a cylinder of which the base is the circle AEB, DFC
the circle opposite to the base, and GH the axis ; from £, any point
in the circumference AEB, let EF be drawn perpendicular to the
plane of the circle AEB : the straight line EF is in the superficies of
the cyhnder.

Let F be the point in which EF meets the
plane DFC opposite to the base ; join EG f xr

and FH ; and let AGHD be the rectangle JD[ — **•

(14. def. 3. Sup.) by the revolution of which
the cylinder ABCD is described.

Now, because GH is at right angles toGA,
the straight line, which by its revolution de-^
scribes the circle AEB^ it is at right angles to
all the straight lines in the- plane of that cir-
cle which meet it in G, and it is therefore
at right angles to the plane of the circle
AEB. «But EF is at right angles to the same
plane ; therefore, EF and GH are parallel
(6. 2. Sup.) and in the same plane. And
since the plane through GHand EF cuts the
parallel planes AEB, DFC, in the straight lines EG and FH, EG is
parallel to FH (14. 2. Sup.). The figure EGHF is therefore a paral-
lelogram, and it has the angle EGH a right angle, therefore it is a rect-
.angle, and is equal to the rectangle AH, because EG is equal to AG.
Therefore, when in theTevolution of the rectangle AH, the straight
line AG coincides with EG, the two rectangles AH and EH will coin,
cide, and the straight line AD will coincide with the straight line EF.
But AD is always in the superficies of the cylinder, for it describes
that superficies ; therefore, EF is also in the superficies of the cyliq-
der. Therefore, &c. Q. E. D,

S14

SUPPLEMENT TO THE ELEMENTS

PROP. XVn. THEOR.

A cylinder and a paraUdopiped htwing equal bases and aJtUudes, are

equal to one another.

Let ABCD be a cylinder, and EF a parallelopiped faaving equal
bases, viz. the circle AGB and the parallelcgram £H, and having also
equal altitudes ; the cylinder ABCD is equal to the parallelopiped EF.

If not, lot them be unequal ; and first, let the cylinder be less thao
the parallelopiped EF ; and from the parallelopiped £F let there be
cut off a part ECl by a plane PQ parallel to NF, equal to the cylinder
ABCD. In the circle AGB inscribe the polygon AGKBLM fliat shall
differ from the circle by a space less than the parallelogram PH (Cor,
1. 4. 1. SupOi and cut off from the parallelogram EH, a part OR equa)
to the polygon AGKBLM. The point R will fall between P and N^
On the polygon AGKBLM let an upright prism AGBCD be constituted
of the same altitude with the cylinder, which will iherefore be less
than the cylinder, because it \^ within it (16. 3. Sup.) ; and if through
the point R a plane RS parallel to NF be made to pass, it will cut ofi*
the parallelopiped ES equal (2. Cor. 8. 3. Sup.) to the prism AGBC,
because its base is equal to that of the prism, and its altitude is the
same. But the prism AGBC is less than the cylinder ABCD, and
the cylinder ABCD is equal to the parallelopiped EQ, by hypothesis ;
therefore, £S is less than EQ, and it is also greater, which is im-
possible. The cylinder ABCD, therefore, is not less than the paral-
lelepiped EF ; and in the same manner, it may be shewn not to be
greater than EF. Therefore they are equal, d. E. D.

OP <3G0METtlT. BOOK Ul^ 914

PROP. XVIII. THEOB.

tf a wne and cylinder fuae the mme ha$B md ^ name niMtHto^ Ae

cane is the third pcfft of the ct/linder.

Let the cone ABCD, and the cylinder BFKG hmre Uie atnne
base, viz. the circle BCD, and the same altitade, ti^. (he perpen.
dicular from the point A upon the plane BCD, Uie. cone ABCD is
the third part of the cylinder BFKG.

If not, let the cone ABCD be the third part of pother cylbder
LMNO, having the same altitude with the cylinder BFKG, but let
the bases BCD and LIM be unequal : and first, let BCD beirreater
thanLIM. *

Then, because the circle BCD is greater than the cirde LIM, jt

Eolygon may be inscribed in BCD, that shall differ from it less than
ilM does.(4. 1. Sup.), and which, therefore, will be greater than
LIM. Let this be the polygon BECFD ; and upon B£CFD,^et there
be constituted the pyramid ABECFD, and the prism BCFKHG.

' Because the polygon BECFD is greater than the circle LIM, the
prism BCFKHG is greater than the cylinder LMNO, for they have
the same altitude, but the prism has the greater base. But the pyra-
mid ABECFD is the third part of th,e prism (15. 8. Sup.) BCFKHG,
therefore it is greater than the third part of the cylinder LMNO«
Now, the cone ABECFD is, by hypothesis,' the third part of the cy-
linder LMNO, therefore the pyramid ABECFD is greater than the
cone ABCD, and it is also less, because it is inscribed in the cone,
which is impossible. Therefore, the cone ABCD is not less than the
^ird part of the cylinder BFKG : And in the same manner, by circum-
scribing a polygon about the circle BCD, it may be shewn that the
cone ABQD is not greater than the third part of the cylinder BFKG^
therefore, it is equal to the third part of that cylinder* Q. E. D,

Sid

SUPPLBMENT TO THE ELEAffiNTS

PROP. XIX. THEOR.

iJTtt hemisphere €md a cone have eqmthaaes and dUiiudesy a series tff
cylinders may be inscribed in the hemisphere j and another series may
he described about the cone^ honing all the same dUUudes with one an*^
vthcTf and such that their sum shcM differ frorA the sum of the head*
sphere^ and the cone^ by a solid less than any given sclid^

Let ADB be a semicircle of which the centre is C, afid let Ct) b^
at right angles to AB ; let DB and DA be squares described on DC,
iiXKSf CE, and let the figure thus constructed revolve about DC: then,
the sector BCD, which is the half of the semicircle ADB, will de-
licribd a hemisphere having G for its centre (7. def. 3. Sup.), and the
ttiangle CDE will describe a cone, having its vertex to C, and having
for its base the circle (11. def. 3. Sup.) described by DE, equal to
that described by BC, which is the base of the hemisphere. Let Vt
be any given solid. A series of cylinders may be inscribed in the
hemisphere ADB, and another described about the cone ECt, so that
their sum shall differ from the sum of the hemisphere and the cone,
by a solid less thah the solid W.

l7pon the base of the hemisphere let a cylinder be constituted
fd()u8J to W, and let its altitude be CX. Divide CD into such a num<*
ber of equal parts, that each of them shall be less than CX ; let these
be CH, rtG, GF, and FD. Through the points F, G, H, draw
FN> GO, HP parallel to CB, meeting the circle in the points K, L
and M ; and the straight line CE in the points Q, R and S. From
the points K, L, M draw Kf, Lg, Mb, perpendicular to GO, HP and
CB ; and from Q, R, and S, draw Qq, Rr, Ss, perpendicular to the
same lines. It is evident, that the figure being thus constructed, if the
whole revolve about CD, the rectangles Ff, Gg, Hh will describe cy-
linders (14. def. 3. Sup.) that will be circumscribed by the hemi-
sphere BDA ; and the rectangles DN, Fq, &r, Hs, will also describe
cylinders that will circumscribe the cone ICE. Now, it may be de*
monstrated, as was done of the prisms inscribed in a pyramid (13. 3.
Sup.), that the sum of all the cylinders described within the hemi«
sphere, is exceeded by the hemisphere by a solid less than the cylinder

6f GEOMETRY. BOOK tlfi 81f

genbtatdd by the rectangle HB, that is, by a solid less than W, for tbcf
cylinder generated by HB is less than W. In the same manner^ it
inay be demonstrated, that the sum of^ the cylinders circumscribing
the cone ICE is greater than the cone by a solid less than the cylinder
generated by the rectangle DN, that is, by a solid less than W. There^
♦fore, since the Sum of the cylinders inscribed in the hemidphefeV
together with a solid less than W, is equal to the hemisphere ; and;
since the sunl of the cylinders described about the cone is equal id
the cone together with a solid less than W ; adding equals to equals;
the sum of all these cylinders, together with a solid less than W, is
equal to the sum of the hemisphere and (he cone together with a
solid less than W. Therefore, the difierence between the whole o^
the cylinders and the suni of the hemisphere and the cone^ is equat
to the difference dftwo Solids, which are each of them less than W ;
but this difierence must also be less than W, therefore the difierence?
between the two series of cylinders and the isum of the hemisphere
tod cone is less than the given solid W. Q. £. D^

PROP. XX.

i^he same things teihg supposed ^as in the lasi prapostiionf the suni of
all the cylinders inscribed in the hemispJiert, and described about the
cone, is equal to a cylinder ^ having the same base and aUitude with
the hemisphere

liet the ^giire I)C6 be constructed im before, send snpp6seG( id
devolve about CD ; the cylyiders inscribed in the- hemisphere, that is;
ihe cylinders described by the revoTirtion of thd rectangles Hh, Gg;
Ff, together with those described about the coire, that is, the cylinder^
described by the revolution of the rectangles Hs, Gr, Fq, anid DN arcf
^qual to the eylinder described by the revolution of the rectanglef

Let L be ihe point in whicli GO rheetti the circle ADB, then, because
CGL is a right atigle if CL be joined, the circles described with thtf
distances CG and GL ate* equal to. the circle described with the dis^
iance CL (2. Cor. 6. 1. Slip.) or GO ; now, CG is equal to GR, be>
&aase CD is equal to D£, and therefore a^o, the tirctesT described
with the distances GR and GL are together equal to the circle desibrib^
ed with the disti^nce GO, that is, the circles described by the revo-
lutioh of GR and 6L about the point G, are together equaf ttf thcf cir-
cle described by the revcrtution of GO about the safhe point G ; there*-'
fore^lso, the cylinders that stand upon the two first of thesfe circles^
fiatihg the common altitudes GH, are equal to the CyKnder whichr
stands on the refnaining circle, and which has the same altitude GU.
The cylinders described by the revolution of the rectangles Gfg,- iinrf
Or are therefore equal to the Cylinder described by the rectangle GP.-
And as the same may be shewn of all the rest, therefore the cylin*^
4eFs described by the rectangles Hh, Gg, Ff, and by the red

218 SUPPLEMENT TO THE ELEMENTS, &c.

Hfl, 6r, Fq, PN, are together equal to the cylinder described by DB^
that is, to the cylinder having the same base and altitude with the
hemisphere. Q. E* D^

PROP. XXL

Every sphere is two4hirds of the circumscribing cylinder.

Let the figure be constructed as in the two last propositions, and if
die hemisphere described by BDC be not equal to two-thirds of the'
eylinder described by Bt), let it be greater by the solid W. Then,*
as the cone described by CDE is one-third of the cylinder (18. 3. Sup.-!
ilescribed by BD, the cone and the hemisphere together will exceed^

the cylinder by W. Bui that .cylinder is' equal to thesum of air the;
cylinders described by the rectangles Hh, Gg, Ff, Hs, Gr, Fq, DN
(20. 3. Sup.) ; therefore the hemisphere and the cone added'together'
exceed the sum of all these cylinders by the given solid W, whicdi
is absurd ; for it has been shewn (19. 3^ Sup.), that the hemisphere
and the cone together differ from the sum of the cylinders by a solid'
less than W. The hemisphere is therefore equal to two-thirds of the
cylinder described by the rectangle BD; and therefore the whole"
sphere is equal to two-thirds of the cylinder described by twice the^
rectangle BD, that is, to two-thirds of the circumscribing cylinder
Q« E. D«

BSD OF THE SUPPLEME17T' TO THE SLEMENTSv^

ELEMENTS

PLANE AND SPHERICAL

TRIGONOMETRY.

ELEMENTS

PLANE TRIGONOMETRY.

TsiflONOMETKY is the application of Arithmetic to Geoknetry : or,
inore'[Hrecisely, it is the application of numher to express the relations
of the sides and angles of triangles to one another. It therefore ne*
cessarijy supposes the elementary operations-of arithmetic to be mi-
derstood, and it borrows from that science - several of the signs or
characters which peculiarly belong to it. Thus, the product of two
numbers A and B, is either denoted by A.B or AXB ; and the pro-
ducts of two or more into one, or into more than one, as of A+B into
C, or of A+B into C+D, are expres sed thus : ( A+B), C, (A+B)

(C+D), or sometimes thus, A+B X C, and A+B X C+D,
The quotient of one number A, divided by another B, is written

thus, g.

The sign ^ i^ used to signify the square root : Thus \/M is the
(square root of M, or it is a numbe r which, if multiplied in itself, will

produce M. So also, v^M*+N*.is the square root of M'+N', &o.
The elements of Plane Trigonometry, as laid down here, are divided
into three sections : the first explains. the principles; the second de-
livers the rules of calculation ; the third contains the construction of
trigonometrical tables, together with the investigation of some theo*
rems, useful for extending trigonometry to the solution oif the mom
di^cult problems.

SECTION I.

LEMMA I,

4n amg^e at the centre of a cvrde is to four right angles as the arch on
which it stands is to the whole circumference.

Let ABC be an angle at the centre of the circle ACF, standing
on the circumference AC : the angle ABC is to four right angles an
tb^ aich AC to the whole circumference ACF.

PLANE TRIGONOMETRY.

Produce AB till it meet tbe circle in E, and draw DBF perpen-
.dicular to AE.

Then, because ABC, ABD are
•two angles at the centre of the eircle
ACF, the angle ABC is to the angle
ABD as the arch AC to the arch AD,
(33. 6.) ; abd therefore also, the
angle ABC is to four times the angle -^
ABD as the arch AC to four times
the arch AD (4. 5.). .

But ABD is a right angle, and
therefore four time^ the arch AD is
equal to the whole circumference
ACF ; therefore the angle ABC is
tp four right angles as the arch AC to (he whole circumference ACF.

Cor. Equal angles at the centres of different circles stand on arch-
es which have tHe same ratio to their circumferences. For, if the
jQDgle ABC, at the centre of the circles, ACEl, GHK, stand on the
iarches AC, GH, AC is to the whoie circumference of the circle ACE,
as the angle ABC to four right angles ; and the arch HG is to the
whole circumference of the circle GHK in the same ratio. There,
/ore, dec.

DEFINITIONS.

Iv two straight lines intersect one another in the centre of a circle,
the arch of the circumference intercepted between them is called
>he Measure of the angle which they contain. Thus the arch AC
is the measfire of the angle ABC.

II.

If the circumference of a circle be divided into 360 equal parts, each
of these parts is called a Degree ; and if a degree be divided into
60 equal parts, each of these is called a Minute ; and if a minute be
divided into 60 equal parts, each of them is called a Secand, and
so on. And as many degrees, minutes, seconds, &c. as are in any
arch, so many degrees, minutes, seconds, dec. are said to be in the
angle measured by that arch.

Cor. 1. Any arch is to the whole circiiinference ojT which it is a
part, as the number of degrees^ and parts of a degree contained in it
is to the number 360. And any angle is to four right angles ad the
number of degrees and parts of a degree in.the arch, which is the
measure of^hat angle, is tp 360,

PLANE f RIGONQMETRt.

S»$

Cor. 2.. Hence also, the arches which measure ^ the same ang^e,
whatever be the radii with which they are described, contaiki the^tne'
iiumbeT of degrees, and parts of a degree. For the number of ile«'
grees abd parts of a degree contained in each of these arches has the*
same ratio to the number 360, that the angle which they measordf"
has to four right angles (Cor. Lem. I.).

The degrees, minutes, seconds, &c. contained in any arch or angle,*
are usually written as in this example, 49^. 36'. 24". 42'" ; that is,'
49 degrees, 36 minutes, 24 secondsj and 42 thirds.

III.

Two angles, which are together etjual tO' two right angles, or two*
* arches which are together equal to a semicircle, are called the Sup^
plemenls of one another.

IV.

A straight line CD drawn through C, one of the extremiUes of ihe^
arch AC, perpendicular to the di- ^ _.

ameter passing through the other — "t^-- ■ ■ * ^

extremity A, is called the Sine of
the arch AC, or of the angle ABCj
of which AC is the measure.

Cor. 1. The sine of a quadrant, or ]g«j
of a right angle, is equal to the ra^
dius.

CoR. 2. The sine of an arch is half
the chord of twice that arch : this'
is evident by producing the sine
of any arch till it cut the circum-
ference.

The segment £)A of the diameter passing through A, one extremity of
the arch AC, between the sine CD and the point A, is called the'
Versed sine of the arch AC, or of the angle ABC.

VI.

A straight line AE touching the circle at A, one extremity of the afdh^
AC, and meeting the diameter BCj which passes through C the*
other extremity, is Called the Tangent of the arch AC, or of the^
angle ABC.

Cor. The tangent of half a right angle is equal to the radius.-

The straight line BE, between the centre and the extremity of tlie'
tangent A £ is called the Secant of the arch AC, or of the angle ABC. >

€oR. to Def. 4, 6, 7, the sine, tangent and secant of any angle ABC,
are likewise the sine, tangent^ and secant of its supplements CBF«^

tnu

PLANE TRlGONGMfiTRt.

It is manifest, from Def. 4. that CD ts the sine of the angle CFF^.
Let CB be produced till it meet the circle again in I ; and it is alscr
manifest, that AE is the tangent^ and BE the secant, of the angle
ABI, or CBF, frtm Def. 6, 7.

Cob. to Def< 4, 5, 6, 7. The sine, versed sine, tangent, and secant of ftf#
arch, which is the measure of any ^

given angle ABC, is to the sine,- , "^ - ^

versed sine, tangent and secant, of
any other arch which is the mea-
sure of the same angle, as the ^^
radius of. the first arch is to the
tadius.of the second.

om:

Let AC, MN be measures of the afi' £
gle ABC, according to Def. I. ;
CB the sine, DA the versed sine. AE the tangent, andBE the se^.
cant of the arch AC, according to Def. 4,^5, 6, 7 ; NO the sine, OM
the versed sine, MP the tangenf, and BP the secant of the arch MN,'
Hccording to the same definitions. Since CD, NO, A£, M P are pa-
tallel, CD : NO : : rad. CB : rad. NB, and AE : MP : : rad. AB t
rad. BM, also Bfi : BP : : AB : BM ; likewise becaus^BC : BD : ^
BN : BO, that is, BA : BE) : : BM : BO, by conversion and alter.^
nation, AD : MO : : AB : MB^ Hence the corollary is manifest.^
And therefore, if tables be constructed, exhibiting in numbers the
sinesy tangents, secants, and versed sines of certain angles to tf
given radius, they will exhibit theratios of the sines, tangents, 6cc^
of the same angles to any radius whatsoever.

In such tables, which are called Trigonometrical Tables, the radius^
is either supposed 1, or some in the series 10, 100, lOOO, &c^
The use and construction ofthese tables are about to be explained.'

Vllt

The difference between any angle and a right angle, or between' an/
arch and a quadrant, is called ^ ._.

the CompZemenf of that angle, or - -^'

of that arch. Thus, if BH be
perpendicular to AB, the angle
CBH is the complement of the
angle ABC, and the arch HC
the complement of AC*; also -pi
the complement of the obtuse
angle FBC is the angle HBC,
its excess above a right angle ;
and the complement of the arch

PLANE TRIGONOMETRT-

IX.

The sine, tatigent, or secant of the complement of any angle is called
the Cosine, Cotangent, or Cosecant of that angle. Thus, let GL or
DB, which is cqtial to CL, he the sine of the angle OBH ; HK the
tangent, and BK the secant of the same angle : CL or BD is the

. cosine, HK the cotangent, and BK the cosecant of the «uigle ABC

"CoK. L The radius is a mean proportional between the tangent andl
the cotangent of any angle ABC ; that is, tan. ABC Xcot. ABC=s

For, since HK, BA are parallel, the angles HBK, ABC are €qual,

and KHB, BA£ are right angles ; therefore the triangles BAE,

/ KHB are similar, and therefore AE is to AB, as BH or BA to HK^

CoH. 2. The radius is a mean proportional between the cosine and
secant of any angle ABC ; or
cos. ABCXsec. ABC=R''.

"^ince CD, A£ are parallel, BD is to BC «r BA, as BA to BE.

PROP,, L

in a right angled plane triangle, as the h$fp&tenuse lo-eitherof the
sides, so the radius to the sine of the angle opposite to thcU side ; and
-as either of the sides is to the other side, eo is the radius to the tan*
£ettl of the amgle opposite to that side.

Let ABC be a right a&gled plane triangle, of which BC is the hy^
|>otenu8e. From £e centre C, with any radius CD, describe \\n^
:arch DE ; draw DF at right angles to CS, and from £ draw £0
touching the circle in £, and meeting CB in G ," DF is the m^ a^id
£G the tangeot of the arch DE, or of the angle C.

The two triangles DFC, BAC are equiangular, because the angles^
DFC, BAC are right angles, and
the angle at C is common. There-
fore, CB : BA : : CD : DF ; but
CD is tho radius, and DF the
-sine of the angle C, (Def. 4.) ;
therefore CB : BA : : R : sin.
CJ.

Also, because EG touches the
circle in E; CEG is a right angle,
and therefore equal to the angle
BAC ; and since the angle at C is common t^ the triangles CBA>
CJGE, these triangles are equiangular, wherefore CA : AB ; : CE ^
EG ; but CE is the radius, and EG the tangent of the angle C ;
therefore, CA : AB : : R : tan. C.

. Cor. 1. As the radius to the secant of the angle C, so is the side
adjacent to that «ngle to the hypotenuse. For CG is the secant of

:2Q

226

PLANE TRIGONOMETRY,

thiB angle G (def. 7.), and the triangles CGE, CBA being equiangu^
lar, CA : CB : : CE : CG, that is, CA : CB : : R : sec. C.

Cor. 2. If the analogies in this proposition, and in the above co-'

roUary be arithmetically expressed, making the radius =1, they give

^ AB AB ^ BC -, ^

am. C=^7^ ; tan. C — -r-ri, sec. C =-r7^. Also, since sin. O =

AC'

AC*

cos B, because B is the complement of C, cos. B ^^9 &nd for the

r. A,C

same reason, cos. C =:^?^*

CoR. 8. In every triangle, if a perpendicular be drawn from any
of the angles on the opposite side, the seg-
ments of that side are to one another as
the tangents of the parts into which the
opposite angle i^ divided by the perpendi-
cular. For, if in the triangle ABC, AD
be drawn perpendicular to the base BC,
each of the triangles CAD, ABD being
right angled, AD : DC : : R : tan. CAD,
•and AD : DB : : R : tan. DAB ; therefore,
•ex aequo, DC : DB^: : tan. CAD : tan: BAD.

SCHOLIUM.

The proposition, just demonstrated, is most easily remembered, by
Btating it thus : If in a right angled triangle the hypotenuse be made
the radius, the sides become the sines of the opposite angles ; and if
one of the sides be made the radius, the other side becomes the tan-
gent of the opposite angle, and the hypotenuse the secant of it.

PROP. II.

The sides of a plane triangle are to one another as the sines of the

opposite angles*

From A any angle in the triangle ABC,
let AD be drawn perpendiculat to BC.
And because the triangle ABD is right
angled at D, AB : AD ; : R : sin. B ; and
for the same reason, AC : AD : : R :
sin. C, and inversely, AD : AC : : sin.
C 2 R ; therefore, ex sequo inversely,
AB : AC : : sin. C : sin. B. In the same
manner it may be demonstrated, that AB : |4
BC : : sin. C : sin. A. Therefore, &c. ^
Q. E. D.

PLANE TRIGONOMETRY.

227

PROP. III.

TJie sum of the sines of any ttoo arches of a circle, is to the difference
of their sines, as the tangent of half the sum of the arches to the tan*
gent cf half their difference.

Let AB, AC be two arches of a circle ABCD ; let E be the eent^e^
and AEG the diameter which passes through A ; sio, AC+sin. AB :.
sin AC-^sin. AB : : tan. ^ (AC+AB) : tan i (AC— AB).

Draw BF parallel to AG, meeting the circle again in F. Draw BEL
and CL perpendicular to AE, and they will be the sines of the archea
AB and AG ; produce CL till it meet the circle again in D ; join DF,;
FC, DE, EB, EC, Da

. Now, since EL from the centre is perpendicular to CD, it bisect*
the line CD in L and the arch CAD
in A r DL is therefore equal to
LB, or to the sine of the arcn AC ;
and BH or LK being the sine of

AB, DK is the sum of the sines
of the arches AC and AB, and CK
is the diflTerence of their sines ; 6j
DAB also is the sum of the arch-
es AC and AB, because^ AD is
equal to AC, and BC is their dif.
ference. Now, in the triangle
DFC, because FK is perpendicu-
lar to DC, (3. cor. 1.) DK : KC
: : tan. DFK : tan. CFK ; but tan. DFK:=tan. ^ arc. BD, because the
angle DFK (20. 3.) is the half of DEB, and therefore measured by
half the arch DB. For the same reason, tan. CFK=tan. ^ arc. BC ;
and consequently, DK : KC : : tan ^ arc. BD : tan. ^ arc. BC. But
DK is the sum of the sines of the arches AB and AC ;"and KC is the
difference of the sines; also BD is the sum of the arches AB and

AC, and BC the difference of those arches. Therefore, &c. Q. E. D.

Cor. 1. Because EL is the cosine of AC, and EH of AB, FK ia
the sum of these cosines, and KB their difference ; for FK=^ FB-f-
EL=EH-f EL, and KB=LH=EH— EL. Now, FK : KB f : tan.
FDK : tan. BDK; and tan. DFK=^cotan. FDK, because DFK is the
complement of FDK ; therefore, FK : KB : : cotan. DFK : tan. BDK,
that is, FK : KB : : cotan. ^ arc. DB : tAn. ^ arc. BC. The sum of
the cosines of two arches is therefore to the difference of the same
cosines as the cotangent of half the sum of the arches to the tangent
of half their difference.

Co». 2. In the right angled triangle FKD, FK : KD : : R : tan.
DFK ; Now FK=cos. AB+cos. AC, KD=sin. AB+sin. AC, and

289 n^ANE TRIGONOMETRT,

tan, DFK = tan. i (AB+AC), therefore cos. AB+coa. AC: sfu.
AB + sin. AC : : R : tan i (AB + AC).

In the same manner, by help of the triangle FKC, it may be shewn
Ifaat eos. AB+ cos. AC : sin*^ AC — sin. AB r : R : tan. i(AC — AB)»

Cor. 3. if the two arches AB and AC be together equal tO'
90^, the tangent of half their sum, that is, of 45^, is equal to the ra-
dius. And the arch BC being the excess of DC above DB, or above
90^, the half of the arch BC will be equal to the excess of the half
ef DC above the half of DB, that is, to the excess ef AC above-
45*^ ; therefore, when the sum of two arches is dO°, the sum of the
•ines of those arches is to their difference as the radius to the tan-
gent of the difference between either of them and 45^.^

PROP. IV..

The 9um of ant^ two sides of a triangle is to their difference, 09 the torn-
gent ^f hdf the sum of the angles opposite to those sides, t^ the io^-
gent of half their difference.

Let ABC be any plane triangle ;
CA+ AB : CA— AB : ; tan \ (B+C) r tan. J (B— C).
For (2.) CA : AB: : sin. B : sjn C ;
and therefore (E» 5.)

CA+AB : CA- AB : : sin. B+sin. C : sin. B-sin. C.
But, by the last, sin. B+sin. C: sin. B — sin. C : :
tan I (B+C) : tan J (B— C) ; therefore also, (11.6.)
CA+AB : CA— AB : : tan ^ (B+C) : tan J (B— C).
Q. E.D,

Otherwise, without the 3d.

Let ABC be a triangle ; the sum of AB and AC any two sides, r9
to the difference of AB and AC as the tangent of half the sum of tl^
angles. ACB and ABC, to the tangent of half their difference..

PLANE TRIGONOMETRY. 239

About the centre A with the radius AB, the greater of the two sides,
describe a circle meeting BC produced in D, and AC produced in
E and F. Join DA, EB, FB : and draw FG parallel to CB, meeting
EBinG.

Because the exterior angle EAB is equal to the two interior ABC,
ACB, (3^. 1.) : and the angle EFB, at the circumference is equal to
half the angle EAB at the centre (20. 3.) ; therefore EFB is half
the sum of the angles opposite to the sides AB and AC.

Again, the exterior angle ACB is equal to the two interior CAD,
ADC, and therefore CAD is the difference of the angles ACB, ADC,
that is, of ACB, ABC, for ABC is equal to ADC. Wherefore also
DBF, which is the half of CAD, or BFG, which is equal to DBF,
is half the difference of the angles opposite to the sides AB, AC.

Now because the angle FBE, in a semicircle is a right angle, BE isr
the taogentof the angle EFB, and BG the tangent of the angle BFG
to the radius FB ; and BE is therefore to BG as the tangent of half
the sum of the angles ACB, ABC to the tangent of half their differ-
ence. Also CE is the sum of the sides of the triangle ABC, and CF
their difference ; and because QC iaparallel to FG, CE : CF : : BE
: BG, (2. 6.) that is, the sum of the two sides of the triangle ABC is
to their difference as the tangent of half the sum of the angles oppo-
site to those sides to the tangent of half their difference. Ct. E. D.

PROP. V. THEOR.

If a perpendicular be dravmfrom any angle of a triangle to the oppo*
site stde^ or base ; the sum of the segments of the base is to the sum
of, the other two sides of the triangle as the difference of those sides
to the difference of the segments of the base.

For (K. 6.), the rectangle under the sum and difference of the seg.
meats of the base is equal to the rectangle under tbe sum and differ-

2S0

PLANE TRIGONOMETRY,

eDce of the sides, and therefore (16. 6.) the sum of the segments of
the base is to the sum of the sides as the difference of the sides to
the difference of the segments of the base. Q. £. D.

PROP. VI. THEOR.

In any triangle, twice the rectangle contained by any ttto sides is to the
difference between the sum of the squares of those sides ^ and the square
of the base, as the radius to the cosine of the angle included by the
two sides.

Let ABC be any triangle, 2AB.BC is
to the difference between AB^+BC^ and
AC^ as radius to cos. B.

From A draw AD perpendicular to
BC, and (12. and 13. 2.) the difference
between the sum of the squares of AB
and BC, and the square on AC is equal
to2BC.BD.

But BC.BA : BC.BD : : BA : BD : : _ _

R:cos. B, therefore also 2BC.BA : B D G

2BC.BD : : R : cos. B. Now 2BC.BD is the difference between AB*^
+BC' and AC*, therefore twice
the re.ctangle AB.BC is to the dif-
ference between AB'+BC^ and
AC^ as radius to the cosine of B.
Wherefore, &c.' Q. E. D.,

CoR. If the raduis=l,BD=BA
Xcos. B, (1.), and2BC.BAXcos.
B*=2BC.BD, and therefore when
B is acute, 2BC.BA Xcos. B= BC^
+BA'— A'C^ and adding AC^* to"
holh; AC^+2 cos. BxBC.BA =
BC*+BA" ; and taking 2 cos. BX BC.BA from both, AC»=BC^— 2
COS. BXBC.BA+BaV Wherefore AC=^(BC^— 2 cos. BxBC.
BA+BA»).

If B is an obtuse angle, it is shewn in the same way that AC =

V (BC'*+2 cosTBXBC.BA+BA^.

PROP. VIL

Four times the rectangle contained by any two sides of a triangle, is to
the rectangle contained by tv>o straight lines, of which one is the base
or third side of the triangle increased by the difference of the two sides,
and the other the base diminished by the difference of the same sides,
as the square of the radius to the square of the sine of half the angle
included between the two sides of the triangle*

LetA6C1)e a triangle of which BC is the base, and AB the
greater of the two sides ; 4AB.AC : (BC+(AB— AC)) X(BC— (AB*
— AC))::R»:(8in.iBAC)».

PLANE TRIGONOMETRY.

881

Produce the side AC to D, sq that AD=AB ; join BD, and draw

A£, CF at right angles to it ; from the centre C with the radius CD
describe the semicircle GDH, ciUting BD in K, BG in G, and meet^'
ing BC produced in H.

It is plain that CD is the difference of the sides, and therefore that *
BH is the base increased, and BG the base diminished by the differ-
ence of the sides ; it is also evident, because the triangle BAD iB
isosceles, that DE is the half of BD, and DF is the half of DK, where
fore DE— DF= the half of BD— DK, (6. 5.) that is, EF=i BK.^
And because AE is drawn parallel to CF, a side of the triangle CFJt>^
AC : AD : : EF : ED, (2. 6.) ; and rectangles of the same, altitude
being as their bases ACAD : AD'' : : EF.ED : ED.^ (1. 6.), and
therefore 4AC.AD : AD^ : ; 4EF.ED : ED', or alternately, 4 AC. AD
: 4EF.ED : : AD» : ED^

But since 4EF=2BK,4EF.ED=2BK.ED=2ED.BK«=DB.BK
=HB.BG ; therefore, 4AC.AD : DB.BK : : AD^ : ED». Now AD
: ED : : R : sin. EAC=sin f BAG (1. Trig.) and AD^ : ED» : : R» :
(sin. i BACy : therefore, (11. 5.) 4AC.AD : HB.BG : : R»: (sin. f

,» BAC)^ or since AB= AD, 4AC.AB : HB.BG : : R» : (sin. ^ BAC)».

'Now 4AC.AB is four times the rectangle contained by the sides of
the triangle ; HB.BG is that contained by BC+(AB— AC) and BC
— (AB— AC)- Th erefo re, &c. Q. E. D.

CoK. Hence 2 v^AOAD : y/HB.BG : : R : sin ^ BAC.

PROP. VIII.

Four times the rectangle contained by any two sides of a triangle, is to
the rectangle contained by two straight lines, of which one is the sum
- of those sides increased by the base of the triangle, and the other the
sum of the same sides diminished by the base, as the square of the
radius to the square of the cosine of half the angle included between
the two sides of the triangle.

Let ABC be a triangle, of which BC is the base, and AB the great,
er ofthe other two sides, 4AB. AC : (AB+AC+BC) (AB+AC— BC)
: : R^ : (cos. i BAC)^

From the centre C, with the radius CB, describe the circle BLM,
meeting Afc, produced, in L and M. Produce AL to N, so that AN
=AB ; let AD=AB ; draw AB perpendicular to BD ; join BN, and

PLANE TRIGONOMETRY.

let it meet the circle agaio in P^ let CO be perpendicular to BN ;
and let it meet AE in R.

It 18 evident that MN=AB+AC+BC ; and that LN=:AB+AC
— BC. Now, beteause BD is bisected in E, (3. 3.) and DN in A,
BN is parallel to AE, and is therefore perpendicular to BD, and the
triangles DAE, DNB are equiangular ; wherefore, since DN =
2AD.BN=2AE, and BP=2BO=2RE ; also PN=2AR.

But because the triangles ARC and AED are equiangular, AC :
AD : : AR : AE, and because rectangles of the same altitude are as

their bases (1. 6.)» ACAD : AD« : : Aft.Afi : AE^ and alternately
ACAD : AR.AE : : AD^ : AE^ and 4AC.AD : 4AR.AE : : AD» :
AE^ But 4AR.AE=2ARX2AE=NP.NB=MN.NL ; therefore
4ACAD : MN.NL : : AD'' : AE«. But AD. : AE : : R : cos. DAE
(l)=cos. i (BAC) : Wherefore 4ACAD : MN.NL : t R» : (cos. J
BACy.

Now 4AC AD is four times the rectangle under the sides AC and
AB, (for AD«=AB), and MN.NL is the rectangle under the sum of
the sides increased by the base, and the sum of the sides diminished
by the base. Therefore, &c. Q. E. D.

CoR. L Hence 2v/ACAB : v/MN.NL t : R : cos. i BAC
Cor. 2. Since by Prop. 7. 4ACAB : (BC + (AB— AC)) (BC—
(AB — BC)) : : R^ : (sin*^ BAC)^ ; and as has been now proved
4ACAB : (AB+AC+BC) (AB+AC— BC) : : R^ : (cos. i BAC)» :
therefore, ex aequo, (AB+AC+BC) (AB+AC— BC) : (BC + (AB
—AC)) (BC— (AB— AC)) : : (cos. | BAC)^ : (sin f BAC)^. But
the cosine of any arch is to the siae» as the radius to the tangent of

VhAimTtimGsovasms.

< AB^AC)) (BC ^(AB^AC)) : :.R« : <^ni * BAG)' ; and

V^ (Ae+A€+B"G) CAB+AC^-^-BO) :

V (BC -PaF- AC)7gC:--( AB— AC)) : « R : tan J^ B AC,

XEMWA JL

J^ tkete he tteo fmeqtud magnitudes^ ^hdf ili^r ,dijfiefeiut3e t$cbied to ialf
their sum is equal to- the greater; and hdlfdktM' d^erenee taieeti
Jrwh half their sum is equal to the less*

Let AB and BO be two luieqfial fiagaitudeS) of which AB is tb»

greater ; suppose AC bisected in D, _

.e^d.A£ji$.quaUoBC. It is manifest A E D B Q

t,^t AC is thce.aum, 9^d;EB the differ.

<ente of the magnitudes. And because AC is bjsec^d in D, AD is
«qual to DC : but A£ is also equal to BC, thetefore^DE is equal to DB^
^d DE or DB is half the difference of the fnagnitudes^ But AB is
tequal to BD and DA, that is,, to half the difference added to half the
sum. ; and BC is ^equal tq the excess of DC, half the sum alcove DB>
half the difier^ee. Therefore, &c, Q.E.D,

I Cox. .Heace^if the «im and the diSecence of two magnitudes he
jgiy^nytben^f^itud^s themselves may. be found ; for to half the siim^
miiiii^f. ihe :differen[ce, and it w;il! give the greater : fr6m half die
mm subtracl half the dif&r^nc^,^ and.it will give tke less.

I^ECTION II.

K)JP THE RULES OF TRIGONOMETttBSAL

CALCULATION.

1!*&E OsNERAL Problem, which Trigonometty proposes to i^8<^vtft
W^ fn amy jkme triangle, of the three sides and tke three an^es^ an§
ihrea h^inggivcM, and one ff these three being, a side, to find isn%^ofihe
xAher^Are^* . ;

The things here said to be givon are understood to be expressed
by their ttvimelocal values : the angles, in degrees^, mimitefii^ •&c« $ and
tiie sides in feeti or ally other known measure^

— . So

S»4

PLANE TRIGONOMETRr.

Tbe reason bf the restriction in this problem to those eases hi
'nvhich at least; one side is given, is evident from this, that by the an-
gles alone beine; given, the magnitudes of the sides are not determin-
ed. Innumerable triangles, equiangular to one another, may exist,
without the sides of any one of them being equal to those of any
other ; though the ratios of their sides to one another will be the same
in them all (4. 6.). If therefore; only the three angles are given,
nothing can be determined of the triangle but the ratios of the sides,
which may be found by trigonometry, as being thoi same with the
ratios of the sines of the opposite angles.

For the conveniency of calculation, it is usual to divide the gene-
ral problem into two ; according as the triangle has^ or has not, one
of the angles a right angle.

PROB. I.

in a right angled triangle, of the three sides, and three angles, any tvfo
being given, besides the right angle, and one of those two being a side,
it is required to find the other three.

It is evident, that when one of the acute angles* of a right angled
triangle is given, the oti^er is given, being the complement of the for*
mer to a right angle ; it is also evident that the sine of any of the
acute angles is the cosine of the other.

This problem admits of several cases, and the solutions, or rules
for calculation, which all depend on the first Preposition, may b^
conveniently exhibited in the form of a table ; where the first column
contains the, things given ; (he second, the things required ; and the
third, the rules or propositions by which they are found. '

GIVEN.

80DSHT.

SOLUTION.

1
2

CB and B^ the
hypotenuse and
angle.

AC.
AB.

•

R : sin B : : CB : AC.
R : <?os B : : CB : AB.

•

AC and C, a

side and one of
the acute angles.

BC. '
AB.

Cos C : R : : AC : BC.

R : tan C : : AC : AB.

3
4

CB and BA,

the hypotenuse
and a side.

C.
AOw

CB.

CB : BA : : R : sin C.
R : cos C : : CB : AC;

5
6

AC and AB,

the two sides.

AC : AB : : R : tan C.
Cos C : R : ; AC : CB.

: 7

PLANE TRIGONOMBTRT. 235

Remarks on the Solutions in the table.

In the second case, when AC and C are given to find the hypote-
nuse BC, a solution may also be obtained by help of the secant, for *
CA : CB : : R : sec. C. ; if, therefore, this proportion be made R :
sec. C : : AC : CB, CB will be found.

In the third case, when the hypotenuse BC and the side AB are
given to find AC, this may be done either as directed in the Table,
or by, t he 47 th of the first ; for since AC^ = BC*— BA«, AC =«

y^BC— BA^ This value oT AC will be easy to calculate by loga- '
rithms, if the quantity BC— BA* be separated into two multipliers,
which may be done; because (C or. 5. 2.), BC— BA'= (BC+BA)

(BC— BA). Therefore AC=V(BC+BA) (BC— BA).

When AC and AB are gtyen, BC may be found from the 47th, as

in the precedii% instance, for BC = ^BAF+AC\. But BA'+AC»
cannot be separated into two multipliers ; and therefore, when BA
and AC are large numbers, this rule is inconv'tinient for computation
by logarithms.' It is best in such cases to seek first for the tangent
of C, by the analogy in the Table, AC : AB : : R : tan C ; but if C
itself is not required, it is sufficient, having found tan. C by this pro.
portion, to take from the ' Trigonometric Tables the cosine that cor-
responds to tan. C,' and then to compute CB from the proportion cos.
€ : R : : AC : CB.

PROBLEM IL

« ■
In an ohlique angled triangle, of the three jsides and three angles, any
three being given, and one of these three being a side, it is required
to find the other three.

This problem has four cases, in each of which the solution depends
on some of the foregoing propositions.

CASE I.

Two angles A and B, and one side AB, of a triangle ABC, bejng
given, to find the other sides.

Because- the angles A and B are gWeo^ O is also giTen, bemg tb»
svpplement of iT+B; aad, (2.)
Sin. C : sin. A t : AB :.BC ; alsoy
Sia. Oim^Bii AB' i AC.

CASE 11.

Two sSter ilB and AC) and the angW B opposite to one of tfren^
ieing given, to- fiiul tbe other angles A and C, and also the other sidir
BC

The stngfe C is foond from this prop<Htiott, A(J : AB : ? 8f»B ?
sin a Also, A:=180^— B^C ; and theiH sin B: sin A r : A€ : CB,.
%y Case 1.

In this ease, the angle C may hare two Tdhies ; (or its sine beid^
Ibund by the proportion above, the angle belonging to*that sine may
either be that which is found in the tables, or it stay berthe supple^
wamdntk (Cor* def. 4.)» This ambiguity, however, does not arise
from any defect in the solution, but from' a circumstaiice^eSsential to»
the problem-, vr2. that whenever AC is less than AB, t^ere are tw»
triangles which have the sides AB, AC, and the dngle at B of ther
same magnitude in each, but which are nevertheless unequal, the-
angle opposite to AB in the (Hie, being the supplement of that which
is opposite to it in the other. The Irutn of this appears by describing
ftom the centre A withahe raiiiu)9 AC» aa arch intersecting 6G in U

PLANE TRI60NOMETRT. 257

and -C' ; then, if AC and AC be drawn, it is evident that the triangles
ABC, ABC have the side -AB and the angle at B common, and the
sides AC and A^' equal, but haver not the remaining, side of fte.one
equal to the remaining side of the other, that is, BC to BC, nor th^it
other angles equal, viz. BCA to BCA, nor BAC to BAC. But in
these triangles the angles ACB, ACB are the supplements of one
another. For the triangle CAC is isosceles, and the angle ACC=
ACC, and therefore, ACB, which* is thf supplement of ACC, is
also the supplement of ACC or ACB; and these two angles, ACB»
ACB are the angles found by the computation above.

From these two angle9> the two angles' BAC, BAC will be found':
the anj^le BAC is the supplement of the two angles ACB, ABC,
(32. 1.), and therefore its sine is the same with the sine of the sum of
ABC and ACB; But BAC is the difference of the angles ACB,
ABC : for it is the difference of the angles AC'C and ABC, because
ACC, that is, ACC is equal to the sum of the angles ABC, BAC,
(32. 1.). Therefore, to find BC, having found C, make sin C : sin
(C+B) : : AB : BC ; and again, sin C : sin (C— B) : : AB : BC.

Thus, when AB is greater than AC^ and C consequently greater
than B, there are two triangles which satisfy the conditions of tlie
question. But wheii AC is greater than AB, the intersections C and
C' fall on dppodite sideVof B, sothat the two triangles have not the
same angle at B common to thiDin^ and the solution ceases to be
ambiguous, the angle required being necessarily less than B, and -
therefore an acute angle.

CASE III.

Two sides AB and AC, and the angle A, between them, beiikg giv*
en to find the other angles B and C, tmd also the side BC.

SOLUTION.

First,'maHe AB+AC : AB- AC : : tan J^ (C+B) : tan i (C-B).
Then, since J (C+B) and i (C — B) are both given, B and C may be
found. For B= i'(C+B)+^(C— B), and e=J (C+B)— i (C-B).
(Lem. 2.)

To find BC.

Having found B, make sin B : sin A : : AC : BC.
^ But BC may also be f ound without seeking for the angte B and C ;

for BC=v/Al^2 cos AXAB.AC+AC*, Prop. 6.

This method of findfiiff BC is eirtremely useful in many geometri-
cal investigations, but it is not very well adapted for computation by
logarithms, because the quantity under the radical sign cannot be se-
parated into simple multipliers. Therefore, when AB and AC are
expressed by iargemutnbdrs^ the other sdution, by finding the angles,
and then computing BC, is prefexiMe*^

2SB

PLANE TRIGONOMETRY.
CASE IV.

The three sides AB, BC, AC, being given, to find the angles A^
B, C.

SOLUTION I.

Take F such that BC : BA+AC : : BA— -AC : F, then F is either
the sum or the difference ofBD, DC, the segments of the base (5.).
If F bo greater thttnBC,F is the sum, and BC the difference of BD,
DC; but, if F bo less than BC, BC is the sum, and F the difference
of BD and DC. In either case, the sum of BD and DC, and their
difference being given, BD and DC are found. (Lem. 2.)

Then, (1. j CA : CD : : R : cos. C ; a^d BA : BD : : R : cos. B ;
wherefore C and B are given, and consequently A.

SOLUTION II.

Let D be the difference of the sides AB, AC. Then (Cor. 7.)
2v/AB7aC : v'(BC+D) (BC-^D) : : R : sin i BAC.

SOLFTION III.

Let S be the sum of the sides BA and AC. Then (1. Cor. 8.)
2v^ARAU t -v/(S+BO) (S— BC) : : R : COS i BAC.

SOLUTION IV. , '

S and D retaining the significations above, (2. Cor. 8.)
v/ (S+BC) (S— BC) : v/(BC+Dy(BC— D) : : R : tan i BAC. -

It may be observed of these four solutions, that the first has the
advantage of being easily remembered, but that the others are rather
more expeditious in calculation. The second solution is preferable to
the third, when the angle sought is less than a right angle ; on the
other hand, the third is preferable to the second, when the angle
sought is greater than^a right angle ; and in extreme cases, that is,
when the angle sought is very acute or very obtusoi this distinctioa

PLANE TRIGONOMETRY.

is very material to be considered. The reason is, that the sines of an-
gles, which are nearly = 90°, or the cosines of angles, which are
nearly = 0, vary very little for a considerable variation in the corres-
ponding angles, as may be seen from looking into the tables of sines
and cosines. The consequence of this is, that when the sine or co-
sine of such an angle is given, (that is, a sine or cosine nearly equal
to the radius,) the angle itself cannot be very accurately found. If,
for instance, the natural sine .9998500 is given, it will be ihimediately
perceived from the tables, that the arch corresponding is between €9°,
and 89° I' ; but it cannot be found true to seconds, becadse the sines
of 89^ and of 89° 1', differ only by 50 (in the two last places,) where-
as the arches themselves differ by 60 seconds. Two arches, there-
fore, that differ by 1'', or even by more than 1", have the same sine
in the tables,. if they fall in the last degree of the quadrant.

The fourth solution, which finds the angle from its tangent, is not
liable to this objection ; nevertheless, when an arch approaches very
nenr to 90, the varations of the tangents become excessive, and are
. too irregular to allow the proportional parts to be found vitb exact-
ness, so that when the angle sought is extremely obtuse, and its half
of -consequence very near to 90, the third solution is the best.

It may always be ^own, whether the angle sought is greater or
less than a right angle by the square, of the side opposite to it being-
greater or less than the s.quares of the other two sides.

SECTION III. .

CONSTRUCTIOPJ OF TRIGONOMETRICAL TABLfiS.

In all the calculations performed by the preceding rules, tables of
sines and tangents are necessarily employed, the construction of
which remains to be explained.

The tables usually contain the sines, <&c. to every minute of the
quadrant from 1' to 90°, and the first thing required to be done, is to
compute the sine of IV or of the least arch in the tables.

1. If ADB be a circle, of which the centre is C, DB, any arch of
tliaf circle, and the arch DBE double of DB ; and if the chords DE,
DB be drawn, also the perpendiculars to them from C, viz. CP,
CG, it has been demonstrated, (8. 1. Sup.) that CG is a mean proper-

• •

PLANE VRIOQNOM£TIUr«

Itooal between AH, half the radius, and AP, the line madeiup of the ra-'
dius and the perpendicular OF. Now OF is the cosine of the arch
BD, and CG the cosine of the halfef BD ; whence the cosine of the
half of any arch BD, of a circle of which the radius ^^ 1, is a mean
proportional between f aqd 1+cos BD. Or, for the greater gene*
rality, supposing A = any arch, cos i- A is a mean proportionaF be-
tween i an d l+.co s A, an d therefore (cos i A)^=V(1 +cos A) or

cos J A=v^ i (1+cos A).

2. From this theorem, (which is the same that is demonstrated
(8. 1. Sup.), only %at it is here expressed trigonometrically,) it is
evident, that if the cosine of any arch be given, the cosine of half
that arch may be found. Let BD, therefore, be equal to 60^, so that
the chord BD = radius, then the cosine or perpendicular CF was
shewn (9. L Sup.) to bo =i, and therefore cos. ^ BD, or cos 30^=3

\/i(l+i')=^\/i==^~o' ^^ the same manner, cos 15°=

v^T(T+cos3p), and cos 7°, 30'=-v/i (1+cos 15°), 4kc. In this
way the cosine of 3°, 45', of 1°, 52', SO", and so on, will be compute
ed, till after twelve bisections of the arch of 60°, the cosine of 52".
44'". 93"". 45^. is found. But from the cosine of an arch its sine
may be found, for if from the square of the radius, that is, from 1,
the square of the cosine be taken away« the remainder is the square
of the sine, and its square, root is the sine ijtself. Thus the sine of
52". 44'". 03"". .45\ is found. . •

3. But it is manifest, that the sines of very small arches are to one
another nearly as the arches themselves. For it has been shewn
that the number of the sides of an equilateral polygon inscribed in a
circle may be so great, that the perimeter of the polygon and the cir-

WLANE TRIGONOMETRY.

241

oumfereDce of the circle may difier by .a line less than any given line,
or, which is the same, may be nearly to one another in the ratio of
equahty. Therefore their like parts will also be nearly in the ratid
of equality, so that the side of the polygon will be to the arch which
tt subtends nearly in the ratio of equality ; and therefore, half the
«ide of the polygon to half the arch subtended by it, that is to say,
the sine of any very small arch will be to the arch itself, nearly in
Uie ratio of equality. I'herefore, if two arches ajre both very small,
the first will be to the second as the sine of the first to the sine of thB
second. Hence, from the sine of 52'. 44", 03'^''. 45\ being found>
the sine of T becomes known ; for, as 52'^ 44"\ 03''". 45'''. to 1, so
is -the sine of the former arch to the sine of the latter. Thus the sine
of r is Ibnnd =:^0.000290d88^

4. The sine 1' being thus founds the sines of 2', of 3', or of any
number of minutes, may be found by the following proposition.

THEOREM.

Let AB, aC, At) be three such arches, that BC the difierence of
%he first and second is equal to CD the difference of the second and
third ; the radius is to the cosine of the common difference BC as the
sine* of AG, the middle arch, to half the sum of the sines of AB and

AD, the extreme arches.

Draw CE to the centre : Tet BF, CG, and DH perpendicular to

AE, be the sines of the arches AB, AC, AD. Join BD, and let it
meet CE in I ; draw IK perpendicular to AE, also BL and IM per*
pendicular to DH. Then, because the
arch BD is bisected in C, EC is at right
angles to BD, and bisects it in I ; also
BI is the sine, and EI the cosine of BC
or CD. And, since BD is bisected in I,
and IM is parallel to BL (2. 6.), LD is
also bisected in M. Now BF is equal to
HL, therefore, BF4-DH=DH+HL=
DL + 2LH === 2LM + 2LH = 2MH or
2KI ; and therefore IK is half the sum
of BF and DH. But because the tri-
angles CGE, IKE are equiangular, CE :
£1 : : CG : IK, and it has been shewn that EI=co8 BC, and IK=i>

(BF+DH) ; therefore R : cos. BC : : sin. AC : i (sin. AB+sin« AD)*
Q. E. D.

Cob. Hence, if the point B coincide with A,
R : cos. BC : : sin. BC : i sin. BD, that is, the radius is to the cosine
of any arch as the sine of the arch is to half the sine of twioe the
arch ; or if any arch=A, f sin. 2A=^sin. AX cos. A, or sin. 2A=2
sin. Ax COS. A.

Therefore also, sin 2'=2' sm I'Xcosl': so that from the sino
•od cosine if one minute the sine of 2' Is found.

am PLANE TRIGONOMETRY.

Again, I', 2^ d\ being three such arches that the difference betireeif
ibe omt and second is the same as between the second and third,
R t eoa r : : sin 2 ; j- (sin r +din 3')t or sin r+sin 8'^2 cos 1'4-
sin 2'» and taking sin 1' from both, sin 9's=^2 cos I'Xsin 2' — sin 1.

In like manner, sin 4' ^2' cos I'Xsin 3' — sin 2^,

sin 5' ==2' cos TXsin 4' — sin 3^,
sin 6'=2' cos r Xsia 5'— sin 4', &c.

Thus a table containing the sines for every minute of the quadrant
nay be computed ; and as the* multiplier, cos 1' remains always ther
same, the calculation is easy.

For computing the sines of arches that difi^r by more than 1', the
method is the same. Let A, A+B, A+2B be three such arches,
then, by this theorem, R : cos B : : sin (A+B) : -J- (sin A+sin (A+
2B)) ; and therefore, making the radius 1,
ain A+sin (A-f-2B)=2 cos BXsin (A+B),

or sin (A+2B)=2 cos BXsin (A+B)— sin A.

By means of these theorems, a table of the sines, and consequently

also of the cosies, of arches of any number of degrees and minutes,.

• ' sin A

from to 90, may be constructed. Then, because tan A=s 1 r

'' cos A

the table oftangents is computed by dividing the sine of any arch by
the cosine of the same arch. When the tangents have been found in
this manner as far as 45^, the tangents for the other half of the quad-
rant may be fonnd more easily by another rule. For the tangent of
an arch above 45^ being the co-tangent of an arch as much under
45^ ;. and the radius being a mean proportional between the tanjs^enl
and coi-taagent of any arch, (1. Cor. def. 9.), it follows, if the difibr*
ence betwei^n any arch and 45^ be called D, that tan (45^ — D) : 1 : ;

1 r tan (45o+D), so that tan (45^+D)= 7 450 _t>\ '

Lastly,, the secants are calculated from (Cor. 2. def. 9.) where it ia

shewn that the radius is a ooean proportional between the cosine and

the secant of any arch, so that if A be any arch, sec A = r*

. COS A

The versed sines are found by subtracting the cosines from the rt*
dins.

5. The preceding Theorentis one of four, which, when arithmeti-
cally expressed, are frequently used in the application of trigonome-«
try to the solution of problems.

Imo, If in the last Theorem, the arch AC='A, the arch BC=Br
and the radius EC=--I, then AD = A+B, and AB=A— B ; and hj
what has just been demonstrated,

I : cos B : :. sin A :. ^ sin (A+B)+i sin (A — B)i

and therefore
sin Axcoa B»i sin (A+E)+ i (A— B).
%dx^ Because BF, IK, DH are parallel, the straight lines BD and
FH are cut proportionally^ and tharefoce FH, the Afievtnce of thr

PLANE TRIGONOMETRY. Sit

straight lines F£ and HE, is biaectad in K ; and therefore, as was
shewn in the last Theorem, KB is half the sum of FE and HEy^lhatis,
of the Cosines of the arches AB and AD. But because of the simi-
tar triangles E6C, EKI, EC : EI : : GB : EK ; now, GG is the
cosine of AC, therefore,

R : cos BC : : cos AG : f oos AD + i cos AB^
or 1 : cos B : : cos A : i cos (A+B) + j- cos (A — B);

and therefore,
cos A X^cos B=J cos (A+B) + ^ cos (A— B)^
Stioj Again, the triangles IDM, OEG are equiangular, for the an-
gles KIM, ElO are equal, being each of them right angles, and there-
fore, taking^away the^angle EIM, the angle DIM is equal to the angle
EIK, that is, to the, angle ECG ; ai^d the angles DMI, CGE are also
equal, being both right angles, and therefore the triangles IDM,
OGE have the sides about their equal angles proportionals, and con-
sequently, EC : CG : : DI : IM ; now, IM is half the difference of
the cosines F£ and EH, therefore,

R : sin AG : : sin : BC : j> cos AB — i cos AD„
or 1 : sin A : : sin B : ^ cos (A — B)— ^ cos (A+B) ;

and also,
sin AX sin B=^ cos (A — B) — ^ cos (A+B).
4to, Lastly, in the same triangles ECG, DIM, EC : EG : : ID :
DM ; now, DM is half the difference of the sines DH and BE,
therefore,

R : cos AC : : sin BC : f sin AD — J sin AB,
or 1 : cos A : : sin B : jl^ sin (A+B)— j^ sin (A+B) ;

and therefore,
cos A Xsia B^f sin ( A+B)^^ sin (A— B).
6. If therefore A and B be any two arches whatseever, the radkie
being supposed 1 ; .

t sin AX cos B^j^ gin (a+B)+} sin (A— B)«
II. eoeAXcosB^icos(A^B)+|oos(A+B).
ill. tAn AXsitt B=f cos (A— B)— ^cos (A+B).
IV, cos AXsin B=| sin (A+B)— | sin (A— B).
From these four Theorems are also deduced ether fimr«

For adding the first and foutth together,
sin AX cos B+cos Axsin B=:sin (A+B).
Also, by taking the fourth from the first
sin Ax cos B — cos AXsin B=sin (A — B).

Again, adding the second and third,
cos AXcos B+sin AXsin B=co8 (A-^B) ;
And, lastly, subtracting the third from the second,
cosAxcqsB — sin AXsin B»cos (A+B).

844 PLANE TRIGONOMETRT-

.7. Agaia, since by .the first of the above theorems, *-

•inAXcosB=i8in(A+B)+^iii(A-B),ifA+B=S,andA-B=I>,

then (Lem. 2.) A = -g""' ^^^ ""2~ ' wherefore sm — ^ X cos

— — - = i sin S+* D. But as S and D n>ay be any arches what-

ever, to preserve the former notation, they may be caHed A and B»

which also express any arches whatever : thus^

. A+B_ A — B . A I 1 . -o
sm— 7. — Xcos-~;r — ^t sm A+t sm B, or

. A+B^^ A — B A I • ■!>

2 sm— ^r — Xcos — jr — =sm A+sm B^

2 2 . *

In the same manner, from Theor. 2 is derived,
2 cos X cos =cos B+cos A. From the 3d,

j^-l-B A. B

2 sin— ^ — Xsin — - — =cos B — cos A ; and from the 4tby

-*• A+B^ . A— B , , ,,

2cos— jr — Xsm — jr — =sm A—sm B»

In all these Tkeorems, the arch B is supposed less than A.

8. Theorems of the same kind with respect to the tangents of
arches may be deduced from the preceding. Because the tangent of
any arch is equal to the sine of the arch divided by its< cosine,

tan ( A+B) = 7*— rW* But it has just been shewn, that

^ cos (A+B)

mn (A+B) = sin A Xcos B+cos A Xsin B^ and that

«09(.A+B) =sco8 A Xcos B — sin A xsin B; therefore, tan (A+B)

sin A Xcos B+cos A Xsin B ,,..,.,,, / *

= V-— K : — irr;- — ri* awd dividmg both the numerator and

cos A X cos B — sm A X sm ^ ^

denominator of this fraction by cos A Xcos B, tan (A + 6)==

-^tan A+tan B _ ,„ . . «x tan A tan B

T"^: T^T — o- In hke manner, tan (A— B)— --t— — r--- — 5.

1 tan A Xtan B ^ ' 1+tan AXtan B

9. If the theorem demonstrated in Prop. 3, be expressed in the
same manor with those above, it gives

sin A+sin B_tan \ (A+B)

sin A-"srnnB""tan J- (A-"B)'
Also by Cor. 1, to the 3d,
cos A+cos B__cot i (A+B)'

cos A— coarR*^tan i (A — Bi*

PLANE TRIGONOMETRY, 245

And by Cor. 2, to the same proposition,

sin A+sin B tan ^ (A+fi) . « . . ^
-r-p s== p 9 or Since R is here supposed =

sin A+sin B
cos A + cos B

=tan i (A+B).

10. In all the preceding theorems, R, th^ radius, is supposed = I,
because in this way the propositions are most concisely expressed,
and are also most readily applied to trigonometrical circulation. But
if it be required to enunciate any of them* geometrically*, the multi«
plier R, which has disappeared, by being made =l,must be restored,
and it will always be evident from inspection in what terms this multi-
plier is wanting. Thus, Theor. 1, 2 sin AXcos B=sin (A+B)-}-
sin (A—B), is a true proposition, taken arithmetically f but taken
geometrically, is absurd, unless we supply the radius as a multiplier
of the terms on the right hand of the sine of equality. It then be-
comes 2 sin A Xcos B=R (sin (A + B) + sin (A—B)) ; or twice the
rectangle under the sine of A, and the cosine of B equal to the rect-
angle under the radius, and the sum of the sines of A-f-B and A—B.

In general, the number oC linear multipliers^ that is, of lines whose
numerical values are multiplied together, must be the same in every
term, otherwise we will compare unlike magnitudes with one another.

The propositions in this section are useful in many of the higher
branches of the Mathematics, and are the foundation of what is call-
ed the Arithmetic of Sines.

ELEMENTS

SPHERICAL

TEIGONOMETRY.

PROP. 1.

If a sphere beeuiUpa plane through the eeiUre^ the seeiwH is m cird$f
kenoimg the same centre mth thesphere^ and equal t^^eireU by the
revohuian ef which the ephere teae deaeribed.

For all the straight lines drawn from the centre to the superficies
of the sphere are equal to the radius of the generating sernicircle,
(Def..7. 3. Sup.). TheriBfore the common section of the spheri-
cal superficies, and of ji plane passing through its centre, is a line,
lying in one plane, and having all its points equally distant from the
centre of the sphere ; therefore it is the circumference of a circle,
(Def. 11. 1.), having for its centre the centre of the sphere, and for
its radius the radius of the sphere, that is, of the semicircle by which
the sphere has been described. It is equal, therefore, to the circle
of which that semicircle was a part. Q. E. D.

DEFINITIONS.
I.

Any circle, which is a section of a sphere by a plane through its
centre, is called a great circle of the sphere.

CoK. All great circles of a sphere are equal ; and any two of them
bisect one another.

They are all equal, having all the same radii, as has just been
shewn ; and any two of them bisect one another, for as they have
the same centre, their common section is a diameter of bdhi and
therefore bisecti both.

SPHERICAL TRIQONOMETAT.

«nr

11.

The pole of a great circle of a sphere is a point in the superficies of
the sphere, from which all straight lines drawn to the circumfer-
ence of the circle are. equal.

III.

A spherical angle is an angle on the superficies of ^ flphere» contain-
ed by the arches of two great circles which intersect one another ;
and is the same with the inclination of the planes of Uiese great
circles.

IV.

A spherical triangle is a figure, upon the superficies of a sphere, com-
prehended by three arches of three great circles, each of whiob
is less than a semicircle.

PROP. II.

The arch of a great circle^ between the pole aaid the cir^mmfaifenee of

another great drde^ u a quefdraxiU

Let ABC be a great circle, and D its pole ; if DC, aa arch of a
. great ciircle, pass through D, and meet ABC i^ C« the arch DC is o
quadrant.

Let the circle, of which CD is an arch^ meet ABC agaia in A, and
let AC be the common section of the jy

planes of these great' circles, which
will pass through E, the centre of the
sphere : Join DA, DC. Because
AD«DC,(Def. 2.), and equal straight
lines, in the same circle, cut ofi* equal xj^
arches, (28. 3.) the arch AD ==._the
arch DC ; but ADC is a semicircle, -»

therefore the arches AD, DC are
each of them quadrants. Q. £. D.

CoR. L If DE be drawn, the angle AED is a right angle ; anci
DE being therefore at right angles to every line it meets with in the
^lane of the circle ABC, is at right angles to that plane (4. 2. Sup^).
Therefore the straight line drawn from the pole of any great circle
to the centre of the sphere is at right angles to the plane of that cir-
cle ; and, conversely, a straight line drawn from the centre of the
sphere perpendicular to the plane of any greater circle, meets the su-
perficies of the sphere in the pole of that circle.

Cor. 2. The circle ABC has two poles, one on each side of it»
pkine, which are the extremities of a diameter of the sphere perpen**
dicular to the plane ABC ; and no other 'points but these two eaiiW
foiea of the circle ABC.

MS SPHERICAL TRIGONOMETRT;

PROP. III.

tf the pcle of a great circle be the same toith the intersection of athbr two
great circles : the arch of the first mentioned circle intercepted be-
tween the other ttoo^ is the measure of the spherical angle which the
«ame two circles make with one another.

' Let the great circles BA, CA on the superficies of a sphere, o^
which the centre is D, intersect one another in A, and let £G be an
arch of another great circle, of which the pole is A ; BC is the mea-
sure of the spherical angle BAG.

Join AD, DB, DC ; since A is the pole of
BC, AB, AC are quadrants, (2.), and the an-
gles ADB, ADC are right angles : therefore
(4. def. 2. Sup.), the angle CDB is the incli-
nation of the planes of the circles AB, AC, and
is (def. 3.) equal to the spherical angle BAC ;
but the arch BC measures the angle BDC,
therefore it also measures the spherical angle
BAC* Q. E. D.

Cob. If two arches of great circles, AB and AC, which intersect
one another in A, be each- of them quadrants, A will be the pole of
the great circle which passes through E and C the extremities of
those arches. For since the arches AB and AC are quadrants, the
angles ADB, ADC are right angles, and AD is therefore perpendicu'
lar to the plane BDC, that is, to the plane of the great circle which
passes through B and C. The point A is therefore (Cor. 1.2.) the
pole of the great circle which passes through B and C.

PROP. VI.

If the planes of two great circles of a sphere be at right angles to one
another i the circumference of each of the circles passes through the
poles of the other ; and if the circumference bfone great circle pass^
through the poles of another ^ the planes of these circles are at right
angles.

Let ACBD, AEBF be two great circles, the planes of which are
right angles to one another, the poles of the circle AEBF are in the
circumference ACBD, and the poles of the circle ACBD in the cir-
cumference AEBF.

From G the centre of the sphere, draw GC in the plane ACBD
perpendicular to AB. Then because GC in the plane ACBD, at

» When in any reference no mention is made of a Book, or of (lie Plant Trigonoroetij,
die Spherical Trigonometiy is meant.

SMlEMCAL TEIGONOMBTWr.

M»

tiglit angles to the plane AEBF,
ia at right angle* to the common
section of the two pkmes, it is
<Def. 2. 2. Sup.) alao at right an-
gles to die plane AEBF, and
dierefore (Cor. 1. 2.) C is the pole
t>f the circle AEBF ; and if CG
be proddced to D, D is the other
pole of the circle AEBP.

In the same raanner,by drawing
OE in the plane AEBF«, perpen-
dicular to AB, and producing it to
F, it has shewn that £ and F are
the poles of the circle ACBD. Therefore, the poles of emch of
these circles are in the circumference of the other.

Again, If O be one of the poles of the circle AEBF, the great
^circle ACBD which passes through C, is at right angles to the circle
AEBF. For, CG being drawn firom the pole to the cmitre t>f tbe
circle AEBF, is at right angles (Cor. 1. 8.) to the plane of that cif*
cle ; and thorfifbro, «very plaii« passitig throtigh CG (17. 2. Sup.) is
at right angles to the plane AEBF ; now, the plane ACBD passes
through CG. l^erefore, &c. Q. £. D.

Cor. 1. If of two great oircles, the first passes through the poles
trf the second, the second also passes through the' poles of the first*
For, if the first passes through the poles of the second, the plane of
the first must be at right angles to the plane of the second, by the
second part of this proposition ; and therefore, by the first part of it,
the circumference of each passes through the poles of the other.

Cob. 2. All greater circles that have a common diameter have
their poles in the circumference of a circle^ the plane of which is
perpendicular to that diameten

PROP. V.

tn isosedes spherical triangles the angles ai the base are efual.

Let ABC be a spherical triangle, havins the side AB equal to the
side AC ; the spherical angles ABC and ACB are equal.

Let C be the centre, of the sphere ;
join PB, DC, DA, and from A on the
straight lines DB, DC, draw the perpen-
diculars A£, AF ; and from the points
E and F draw in the plane DBC the
straight lines £G=, FG perpendicular to
DB and DC, meeting one another in
G : Join AG.

Because DE is at right angles to each
of the straight lines AE, EG, it is at
right angles to the plane AEJG, which

'250 SPHERICAL TRIGONOMETRt.

passes through AE, EG (4. 2. Sup.) ; and therefore, every pldn^
that passes through D£ is at right angles to the plane AEG (17. 2#
Sup.) ; wherefore, the plane DBG is at right angles to the plane
AEG. For the same reason, the plane DBG is at right angles to the
plane AFG, and therefore AG, the common section of the planes
AFG, AEG is at right angles (16. 2. Sup.) to the plane DBC, and
the angles AGE, AGF are consequently right angles.

But since the arch AB is equal to the arch AC, the angle ADB is
equal to the angle ADC. Therefore the triangles ADE, ADF, have
the angles EDA, FDA, equal, as also the angles AED, AFD, which
are right angles ; and they have the side AD common, therefore ^e
other sides are. equal, viz. AE to AF, (26. l.)» and DE to DF. Again,
because the angles AGE, AGF are right angles, the squares on AG
and GE are equal to the square of A£ ; and the squares of AG and
GF to the square of AF. But the squares of AE and AF are equal,
therefore the squares of AG andGE are equal to the squares of AG
and GF, -and taking away the common square of AG, the remaining
squares of GE and GF are equal, and GE is therefore, equal to GF.
Wherefore, in the triangles AFG, AEG^ the side GF is equal tq the
side GE, and AF has been proved to bo equal ta AFi, and the base*
AG is common ; therefore, the angle AFG is equal to the angle AEG
(8. 1 .). But the angle A VG is the angle which the plane ADC makes
with the plane D^C (4. def. 2. Sup.) because FA and FG, which are
drawn in these planes, are at right angles to DF, the common section
of the planes. The angle AFG (3. def. ) is therefore equal to the sphe-
rical angle ACB ; and, for the same reason, the angle AEG is equal
to the spherical angle ABC. But the angles AFG, AEG are equal.
Therefore the spherical angles ACB, ABC are also equal. Qi. E« D.

PROP. VL

If the angles at the base of a spherical triangle he equals the triangle is

isoscelesi

Let ABC be a spherical triangle having the angles ABC, ACB
equal to one another ; the sides AC and AB are also equal.

Let D be the centre of the sphere ; join DB, DC, DA, and from
A on the (straight lines DB, DC, draw the perpendiculars AE, AF f
and from the points E and F, draw in the
plane DBC the straight lines EG, FG
perpendicular to DB and DC, meeting
one another in G ; join AG.

Then, it may be proved, as was done
in the last proposition, that AG is at
right angles to the plane BCD, and that
therefore the angles AGF, AGE are
right angles, and also that the angles
AFG, AEG are equal to the angles
which the planes DAC, DAB make

SPHEBICAL TDGONOMEmr. 9Ht

■

\
I

vith die pine DBC. Eat tir riaw the^>befKal i^esACB, ABC
are eqml. ti>e as^es vkidi due plaacs llAC, DAB «dbe « iUj tAie
pbM DEC are ecmaL '3. del:; aod ti^et^M liK aiii^Jiw
aie also cx^caL TW t naa tf ka AGEL AGF icare iiief«i^.«re tvc^ a<i^iie»
€if d>e <ne eqrai! lo tw ax^es <4' f^ 4iSUer, aikti *jutrT iuir^ aiaci tt^
see AG ODCLaxM, «i«»fi^««r ihtj are eouai, a»i U«e a»de AF m 4i9ii«l

Azaia. hocamat ^^ mn^Mi» ADF, A DE are r^ M;<rM «t F
£. tie 99Bsres of DF vud FA are oomT %» tsie ^tta»i^ V/ i>A^ U«t
i£.ioiaeaq«irea«f DEaiiC DA ; mmt. *3m: «9atoc 'vf AF J» ^i^ml W
t2« fioijare of AE. ^S0tm^jn me aouave ^ DF jt equsb. ^ ^i^e i^fuare
©TDE, aijc :*c ftde DF vs. tue «cie I/E. T*ii««irfwr*- n. -i-t fr-tAfjc'^rt
E*AF, DAE, WecaiaeDF js emnu ^ !>£ aii^ Dl <fmmavi. vuc z.mt
AF equal lo AE. •iie aie^ AZ>F m wubL u ^sue ^ttf^if: Aljh tM3H>
lore Zkiflfb like arcxies ^ C a&c AS. -miuoL astr lue tutsfm^iMi^ M^ tM; ani^
fri^ Al>F and Al 1. ii» «qu& » um: wmmtuv mul iuii VKan^m

av tuxitm

LdBt ABC be a apiieneiL nawgfc mf iwt aois iX HI

tiiaD Vd& isait ant ii!!!l.

l^l Jp ^ tae iguiR c' 2K --

aptiCT^ : ic«L Zl^ JH- !•!- -^

Xnt fiolii: araif^ a: 1 5^ *:n^■-.^
br tnrBt t>i
BDC : air

BDC arc ^assssr 1:2. i_ i?n; ^^
than die thnx Al»7 : maxL tae?B;r»?5 ^
aBT two of toe m ^ "J"'"- ^^ t- ^^ 3:* 1
which
ABand B*:

die tmni A.C.

nefJb«e««rf« of

Let ABC be a
BC, AG are less tlwB die

Let D be the centre of
tiined by three gl
less than (bur right
AC, which are the
four quadrants described
ferenoe^of a i^real circia. W 2 ^

mt SPKEBICAL TRI€K)]!irOIif£TSr.

PROP. IX.

Jb a tphenctH iriangJe the greater angle is opposite to the greater side p

ottd convepseTjf,

Let ABC be a spherical triangle, the greater angle A i» oj^secl
to the greater side B€.

Let the angle BAD be suede
equal to the angle B, and then ^«^

BD, DA will be equal (6.)^ and
tberefoie AD, DO are equal ta
BC ; but AD, DC are greater
thsLa AC (7.), therefore BC is
greater than AC,, that is, the
greater angle A is opposite to the

greater side BC. The converse i^ "^

is demonstrated as Prop. 19. !• -^ ^

Elenu Q. E. D.

PROP. X.

According a» the smt of ttoo of the sides of a spherical tnangh^ w?
greater than a semicircle, equal to it, or less, each of the interior an-^
gles at the hose is greater than the exterior and opposite angle at the^
hue, equal to it, or less; and also the sum of the ttoo interior angles^
at the base greater than ttoo right angles, equdl to two right angles^,
or less than two right angles^

Let ABC be a sphericaF triangle, of which the sides are AB and
BC ; produce any of the two sides as AB, and tho base AC, till they
meet again in D ; then,^ the arch ABD is a semicircle, and the spheri-
eal angles at A and D are equal, because each of them- isthe inelifta-.
lion of the circle ABD to the circle A CD*

1. If AB, BC be equal to -d

a semicircle, that is, to AD, ^.^^ """^^-^s^

BC will be equal to BD, and ^^

therefore (5
or the angle
to the angle

interior angle „, ^ _ _

equal to the eictejrior and op- ^

posite.

2. If AB, BC together be greater than a semicircle, that is> greater
than ABD, BC will be greater than BD ; and therefore (9.), the ai*^
gle D,that is, the angle A, is greater than the angle BCD.

' 8. In the same manner it is shewn, if AB, BC together be lessr
than a semicircle, that the angle A is less than the angle BCD.

3, that is, to AD, ^.^^ """"""^-^

equal to BD, and y"^ \ ^N.

>.) the angle D X \ X

A, win be equal -^V \ /^

BCD, that is, the \^^ X ^y^

;le at the base ^ ^'^.^^^

SPHERICAL TRIGONOMETRT. 368

Now, since the angles BCD^BCA are equal to two right angles, if
the angle A be greater than BCD, A and ACB together will be great-
er than two right angles. If A be equal to BCU, A and ACB togj»-^
ther, will be equal to two right angles ; and if A be less than B(%|
A and ACB will be less than two right angles. Q. £• D.

PROP. XI.

— «

If the angular points of a gpherical triangh de made the poles of three
great drctesy these three circles by their interseetiane wiU form a tti^
angle f which is said to he supplemental to \he former ; and the tw<k
triangles are such, that the sides of tike one are the suppUmeMs of
the arehes which measure the angles cfthe other*

Let ABC be a spherical triangle ; and from the points A, B, and C
as poles, let the great circles F£, £D, DF be described, intersecting
one another in F, D and E ; the sides of the triangle FED are the
supplement of the measures of the angles A, B, C, viz. FE of the
angle BAC, DE of the angle ABC, and DFof the angle ACB: And
again, AC is the supplement of the angle DFE, AB of the angle
FED, and EC of the angle EDF.

Let AB produced meet DE, EF in
G, M ; let AC meet FD, FE in K, L ;
and let BC meet FD, DE in N, H.

Since A is the pole of FE, and the
circle AC passes through A, EF will
pass through the pole of AC ( 1 . Cor. 4.)
and since AC passes through C, the
pole of FD, FD will pass through the
pole of AC ; therefore the pole of AC
18 in the point F, in which the arches
DF, EF intersect each other. In .the
same manner, D is the pole of BC, and
E thepo\e of AB.

And since F, E are the poles of AL, AM, the arches FL and EM
(2.) are quadrants, and FL, EM together, that is,FE and ML toge-
ther, are equal to a semicircle. But since A is the pole of ML, ML
is the measure of the angle BAC (3.), consequently FE is the-sup*
plement of the measure of the angle BAC. In the same manner,
ED, DF are the supplements of the measures of the angles, ABC»
BCA.

Since likewise CN, BH are quadrants, CN and BH together, that
is, NH and BC together, are equal to a semicircle ; and since D is
the pole of NH, NH is the measure of the angle FDE, therefore the
measure of the angle FDE is the supplement of the side BC. In the
same manner, it is shewn that thcmeasuresof the angles DEF, EFD
are the supplenoents of the sides AB, AC in the triangle ABC.
Q. E. D,

354

SPHERICAL TRIGONOMETRT.

PROP. XII.

The three angles of a spherical triangle are greater than ttoo, and les»

tJian six, right angles.

The mes^ures of the angles A, B, O, in the triangle ABC, together
with the three sides of the supplemental triangle DEF, are (it.)
«qual to three semicircles; but the three sides of the triangle. FDE,
are (8.) less than two semicircles; therefore the measures of the^
angles A, B, C are greater than a semicircle ; and hence the angles
A, B, C are greater than two right angles.

And because the interior angles of any triangle, together with th&
exterior, are equal to six right angles, the interior alone are less thaa
six right angfe's. Q. E. D.

PROP. XIII.

If to the circumference of a great circHeffrom a point, in the surface:
of the sphere, tohich is not the pole of that circle, arches of great
circles be drawn ; the greatest of these arches is that which passes
through the pole of the first-mentioned circle, and the supplement of
it is the least ; and of the other arches, that which is nearer to the-
greatest is greater than that which is more remote.

Let " ADB be the circumference of a great circle^ of which the-
pole is H, and let C be any other point ; through C and H let the-
semicircle ACB be drawn meeting the circle ADB in A and B ] and
let the arches CD, CE, CF also be described. From C draw CGI
perpendicular to AB, and then, because the circle AHGB whiclv
passes through H, the pole of the circle ADB, is at right angles to.
ADB, CG is perpendicular to the plane ADB. Join GD, GE, GF.
CA, CD, CE, CF, CB.

Because AB is the diameter of
the circle ADB, and G a point in
it, which is not the centre, (for
the centre is in the point where
the perpendicular from U meets
AB), therefore AG, the part of
the diameter iii which the centre
is, is the greatest (7. 3.), and GB
the least of all the straight lines
that can be drawn from G to the
circumference ; and GD, which is

nearer to AB, is greater than GE, which is more remote. But the
triangles CGA, CGD are right angled at G, and therefore AC^^
AG-+GC^ and DC^=DG^+GC^ ;. but AG»+GC»7'DG»+GC» ;
because AG7DG; therefore AC^7DC^ and AC7DC. And .be.
cause the chord AC is greater than the chord DC, the arch AC i^

SPHERICAL TRIGONOMETRY.

255

gt^ater than the arch DC. In the same manner, since GD is greater
than GE, and GE than GF, it is shewn that CD. is greater than CE,
and CE than CF. Wherefore also the arch CD is greater than the
arch CE, and the arch GE greater than the arch CF, and CF than CB,
that is, of all the arches of greater circles drawn from C to the circum-
ference of the circle ADB, AC which passes through the pole H, is
the greatest, and CB its supplement is the least ; and of the others,,
that which is nearer to AC the greatest, is greater than th&t whicb
is more remote. Q. E. D.

PROP. XIV.

^ aright angled spherical triangle^ the sides containing the right angle
are of the same affection with the angles opposite to them, that is, if
the sides be greater or less than quadrants, the opposite angles tow
he greater or less than right angles, and conversely.

Let ABC be a spherical triangle, right angled at A, any side AB
will be of the same affection with the opposite angle ACB.

Produce the arches AC, AB, till they meet again in D, and bisect
AD in E. Then ACD,'ABD are semicircles, and AE an arch of 90o*
Also, because CAB is by hypothesis a right angle, the plane of the
circle ABD is perpendicular
to the plane of the circle
ACD, so that the pole of
ACD is in ABD, (cor. 1.
4.), and is therefore the point
£• Let EC be an arch of ^
a great circle passing through
E and C.

Then because E is the
pole of "the circle ACD, EC
is a (2.) quadrant, and the
plane of the circle EC (4.)
is at right angles to the plane
of the circle ACD, that is,
the spherical angle ACE is a
right angle ; and therefore, i&
when AB is lesd than AE,
the angle ACB, being less,
than ACEJp is less than a
right angle. But when AB
is greater than AE, the angle
ACB is greater than ACE, or than a right angle. In the same way
may the converse be demonstrated. Therefore, &c. Q. £» D.

AM SPHERICAL TRIGONOlfBTtlY.

PROP. XT.

^ the two sides ofu right angled spherical triangle about the right ati*
glebe of the same affection, the hypotenuse wm be less than a quad-
rant ; and if they be of different affection^ the hypotenuse wiU he
greater than a quadrant.

Let ABC be a right angled spherical triangle ; aocordihg as the
two sides AB, AG are of the same or of different affection, the
hypotenuse BC will be less, or greater than a quadrant*

The construction of the last proposition remaining, bisect the se*
micircle ACD in G, then AG will be an arch of 90"^, and G will be
the pole of the circle A6D.

1. Let AB, AC be each less than 90^. Then, because C is a point
on the surface of the sphere, which is not the pole of the circle ABD«
the arch CGD, which passes through G the pole of ABD is greater
than CE (13.), and C£ greater than CB. But CE is a quadrant, as
was before shewn, therefore CB is less than a quadrant. Thus also
it is proved of the right angled triangle CDB, (right angled at D), in
which each of the sides CD, DB is greater than a quadrant, that the
hypotenuse BC is less than a quadrant.

2. Let AC be less, and AB greater than 90°. Then because CB
falls between CGD and C£, it is greater (12.) than CE, that is, than
a quadrant. Q. £• D.

Cor. 1 . Hence conversely, if the hypotenuse of a right angled tri*
angle be greater or le^s than a quadrant, the sides will be of different
or the same affection.

Cor. 2. Since (14.)/the oblique angles of a right angled spherical
triangle have the same affection with the opposite sides, therefore, ac*
cording as the hypotenuse is greater or less than a quadrant, the ob«
lique angles will be different, or of the same affection.

Cor. 3. Because the sides are of the same affection with the oppo*
site angles, therefore when an angle and the side adjacent are of the
same affection, the hypotenuse is less than a quadrant : and conversely*

PROP. XVL

In any spherical triangle, if the perpendictdar vpoA the base from the
opposite angle fall within the triangle^ the angles at the base are of
the same affection ; and if the perpendicular fM imthout the triangle^
the angles at the base are of different affection.

Let ABC be a spherical triangle, and let the-arch CD be drawn
from C perpendicular to the base AB.

SPHERICAL TRIGONOMETRY.

257

1. Let CD fall withia the triangle ; then» since ADC, BDC ard
iright angled spherical triangles, the angles A, B must each be of the -
same affection with. CD, (14).

2. Let CD fall without the triangle ; then (14.) the angle B is of
the same affection with CD ; and the angle CAD is of the same afiec*
tion with CD ; therefore the angle CAD and B are of the same af-
fection, and the angle CAB and B are therefore of different affections.
Q. E. D.

' Cor. Hence, if the angles A and B be of the same affection, the
perpendicular will fall within the base ; for if it did not, A and B
would be of different affection. And if the Angles A and B be of
different affection, the perpendicular will fall without the triangle ;
for, if it did not, the angles A and B would be of the same affectionj
contrary to the supposition.

PROP. XVIL

If to the hose of a sphericcd triangle a perpendicular be drawn from tk^
opposite anglSf which either falls within the triangle, or is the nearest
of the ty>o that, fall without ; the least of the segtnertts of the base is
adjacent to the least of the sides of the triangle, or to the greatest, ac*
cording as the sum of the sides is less or greater than a semicircle.

Let ABEPbe a gr^at circle of a sphere, H its pole, and GHDaay
circle passing through H, which therefore is perpendicular to the cir«
cle AB£F. Let A and B be two points in the circle ABEF, on oppo*
site sides of the p9int D, and let
D be nearer to A than to B, and
tot C be 'any point in the circle^
GHD between Jl and D. Through
the points A and C» B and C, let
the arches AC and BC be drawn,
and let them be produced till they
meet th^ circle ABEF in the points 6
£ and F, then the arches ACE,
BCF are semicircles. Also ACB,
ACF, CFE, ECB, are four spheri-
cal -triangles continued by arches
of the same circles, and having the
Mmo perpendiculars CD and CG.

258 SPHERICAL TRIGONOMETllY.

1. Now because CE is nearer to the arch CH6 than CB is, CE 10
greater than CB, and therefore CE and CA are greater than CB and
CA, wherefore CB and CA are less than a semicircle ; but because
AD is by supposition less than DB, AC is also less (ban CB, (13*)»
and therefdre in this case, viz. when the perpendicular falls within
the triangle, and when the sum of the sides is less than a semicircle,
the last segment is adjacent to the least side.

2. Again, in the triangle FCA the two sides FC and CA are less
than a semicircle ; for since AC is less than CB, AC and CFare less
than BC and CF. ; Also, AC is less than CF, because it is more re>
mote from CHG than CF is ; therefore in this case also, viz* vfhen
the perpendicular falls withQut the triangle, and when the sum of the
tildes is less than a semicircle, the least segmeot of the base .AD is
adjacent to the least side.

3. But in the triangle FCE the two sides FC and CE ar^ i;reat<»
than a semicircle ; for, since FC is greater than CA,.FC and C£ are
greater than AC and CE. And because AC is less than CB, KC is
greater than CF, and EC is therefore nearer to the perpendicular
CHG than CF is, wherefore EG is the least segment of the ba^e^
and is adjacent to the greater side.

4. In the triangle ECB the two sides EC, CB are greater than a
semicircle ; for, since by supposition CB is greater than CA, EC and
CB are greater than EC and CA. Also, EC ia greater than CB»
wherefore in this case, also, the l«ast segment of the base EG is ad-
jacent to the greatest side of the triangle^ Therefore, when the
sum of the sides is greater than a semicircle, the least segment of
the base is adjacent to the greatest side, whether the perpendicular
fall within or without the triangle : and it has been shewn, that when
the sum of the sides is less than a semicircle, the least segipent of the
base is adjacent to the least of the sides, whether the perpendicular

' fall within or without the triangle. Wherefore, &c. Q, E* D.

PROP. xvni.

In right angled spherical triangles^ the 4ine of either of the M$8 ahomi
tie right angle^ is to the radius of the sphe^re, as the tangent of the
remaining side is to tlie tangent cf the angle opposSle to that stdem

Let ABC be a triangle, having the right angle at A ; and }et AB be
either of the sides, the sine of the side AB will be to the radius, a^
the tangent of the other side AC to the tangent of the angle ABO,
opposite to AC. Let D be the centre of the sphere ; join AP, BDj
CD, and let AF be drawn perpendicular to B.D, which therefore wili
be the sine of the arch AB, and from the point F, let there be di*aw^
in the plane BDC the straight line FE at right angles to BD, meet*
ing DC in E, and let AE be joiiv^d. Since tliere&re die straigM

, SPHERICAL TRIGONOMETRY,

259

Ikie DE is nt right angles to both FA
^d F£> it will also be at right angles
to the. plane AEF (4. 2# Suf),) ; where-
fore, the piano ABD| which passes
through DP, is perpendicular to the
pkne AEP (17. 2. Sup.)^ and the
plane' AEF perpendicular to ABD :
But the plane ACD or AED, is also
perpendicular to the same A6D, be-
cause tb<^ spherical angle BAG is a
jright anffle ; Therefore AE, the com-
mon section of the plsuies A ED, AEF,
is at right angles to the plane ABD,
OS. 2. SupO, and EAF, EAD are

right angles* Therefore AE is the tangent of the arch AC ; and m
the recUiinenl triangle AEF, having a right angle at A* AF is to tbiO
txriios as AEto the tangent of the angle AFE, (I. PI. Tr.) ; but AF
is tl»e Bine of .the arch AB, and AE the tangent of the arch AC ; and
the angle AFE is the inclination of the planes C6D, ABD, (4. def. 2.
Sup.), or is equal to the spherical angle ABC : Therefore the sine of
the arch AB is to the fadius as the tangent of the arch AC to the tan-
gent of the opposite angle ABC. Q. j£. D.

Cob. Since by this proposition, sin AB : R : : tan AC : tan ABC ;
and betcaus^R : cot ABC : : tan. ABC : R(l. Cor. def. 9, PI. Tr.) by
equality, sin AB : cot ABC : : tan AC : R.

PROP. XIX.

In right angled spherical triangles the sine of the hypotenuse is to the
radius as the sine of either side is to the sine of the angle opposite to
that side.

Let the triangle ABC be right angled at A, and let AC be either of
the sides : the sine of the hypotenuse BC will be to the radius as the
sine of the arch AC is to the sine of the angle ABC.

Lei D be the centre of the sphere, and let CE be drawn perpendi-
cular to DB, which will therefore be the sine of the hypotenuse BC ;
and from the point E Ifet there be r«

drawn jn the plane ABD the straight
tine £F perpendicular to DB. and
let CF be joined : then CF will be
at right angles to the plane ABD,
because as was shewn of EA in the
preceding proposition, it is the com- j\
tnon section of two planes DCF,
ECF, each perpendicular to the
plane ADB. Wherefore CFD, CFE
are right angles, and CF is the sine
of thc^archVAC; and in the triangle QF£ having the right angle CFi:

260 SPHERICAL TRIGONOMETRY.

CE is to the radius, as CF tothe sine of the angle CEF(1. PI. Tr.).
But, since CE, FE are at right angles to DEB, which is the common
section of the planes CBD, ABD, the angle CEFis equal to the incli.
nation of these planes, (4. def. 2. Sup.), that is, to the spherical angle
ABC. Therefore the sine of the hypotenuse CB, is to the radius, as
the sine of the side AC to the sine of the opposite angle ABC. Q. E. D.

PROP. XX.

Jn right angled spherical triangles, the cosine of the hypotenuse is to
the radius as the cotangent of either of the angles is to the tangent of
the remaining angle.

Let ARC be a spherical triangle, having a right angle at A, the
cosine of the hypotenuse BC is to the radius as the cotangent of the
angle ABC to the tangent of the an^le ACB.

Describe the circle DE, of which B is the pole, and let it meet AC
in Fy and the circle BC in £> ; and since the circle BD passes through

tlie pole B, of the circle DF, DF must pass through the pole of BD,
,(4.). And since AC is perpendicular to BD, the plane of the circle
AC is perpendicuUir to the plane of the circle BAD, and therefore
AC must also (4.) puss . through 4h« pole of BAD ; wherefore, the
pole of the circle BAD is in the point F, where the circles AC, DE,
intersect* The arches F V, FD are therefore quadrants, and likewise
the arches BD. BE. Therefore, ifi the triangle CEF, right angled
at the point E, CE is the complement of BC, the hypotenuse of the
triangle ABC ; EF is the complement of the arch ED, the measure
of the angle AHC, ?ind FC, the hypotenuse of the triangle CEF, is
the tomplement of AC, and the .arch AD, which is the measure of
the an^le CFE, is the complenient of AB.

But (18.) in the triancrle CEF, sin CE : R : : tan EF : tan ECF,
thai is, in the iriabgle ACi>, Cos BC ; R : ; cot ABC ; tan ACB.
Ci. E. D. •

SPHERICAL TRIGONOMETRY. 261

Cor. Because cos BG t R i: cot ABC : tan ACB, and (Cor. 1.
def. 9. PI. Tr.) cot ACB : R : : R : tan A€B, ex aequo, cot ACB :
• cos BC : : R : cot ABC.

PROP. XXI.

: Inrightahgledsphericcd triangles, thecQjsme of an angle is to the radius
as the tangent of the side adjacent to that angle is to the tangent of the

: ' hypotenuse* : ♦

The same construction remaining ; In the triangle CEF, sin FE :
R : : tan CE : tan CFE (18.) : but sin EF=cos ABC ; tan CE=
cot BC, and tan CFE=^cot AB, therefore cos ABC : R ^: col BC :
cot AB. Now, because (Cor. 1. def. 9, PI. Tr.) cot BC : R : : R :
tan BC, and cot AB : R : : R : tan AB, by equality inversely, cot
fiC : cot AB : : tan AB : BC ; therefore (11. 6.) cos ABC : R : :
tan AB : tan BC. Therefore d&c. Q. E. D.

CoR, 1. From the demonstration it is manifest, that the tangents
of any two arches AB, BC are reciprocally proportional to their

cotanorents. _

Crt». '2. Because cos ABC : R : : tan, AB : tan BO, and K : cos
BC : : tan BC : R, by equality, cos ABC : cot BC : : ta^ AB: R.
That is, the cosine of any of the oblique angles is to tlie cotangent
of the hypotenuse, as the tangent of the side adjacent to the angle is
to the radius.

PROP. XXII.

In right angled spherical triangles, the cosine of either of the sides is
to the radius, as the cosine of the hypotenuse is to the cosine of the
other side.
The saiiie coostructidn remaining : In the triangle CEF, sin CP :

R ; : sin CE : sin CFE, (1».) ; but flin CF===co8 CA, sin CEjrrcos

BC, ana sin CFE = cos AB ; therefore cos CA s R s s cos BC ; cw

AB. ft.E. D.

8^Hi»I0AI> T^IQONOMEptT.
PROP. XXIII,

In right angled spherical triangles^ the cosine qf either of the sides iaio
the radius t as the cosme of the angle opposite to that side is to the
sine qf the other angle.

Th« Slime coostroctioQ regiainii^ : in the triaogle C£F, sin OF :
R : : Bin EF : ^in EOF, (19.) ; bm sin CF==^cos CA, sin fiF^ccos
ABC, nnd sin ECF=sin .BCA : therefore, cos CA : R : ; cos ABC :
sin BCA. <a. £. a

PROP. XXIT.

fn spherieat triangles, whether right angled or obltbite angled, the sines
. of the sides are proportional to the sines of the angles opposite to them*

First, let ABC be a right angled triangle, having a riglft angle Ut
A ; therefore, (19.) the sine of the hypotenuse BC is to the radius,
' (or the sine of the right angle at A), ri
as the sine of the side AC to the
sine of the angle B. And, in like
manner, the sine of BC is to the . ^^^^

sine of the angle A, as the sine of ' ^ -^*^B

AB to the sine of the angle C ;
wherefore (11. 5.) the sine of the
side AC is to the sine of the angle pX
B, as the side of AB to the sine of
the angle C.

Secondly, Let ABC be an oblique angled triangle, the sine of any
of the sides BC will be to the sine of any of the other two AC, as tha
side of the angle A opposite to BC, is to the sine of the angle. B op.
posite to AC. Through the point C, let there be drawn an arch of a
jr^ttt circle CD perpendicular to AB ; and in the right angled trian-

B D

S^SLS P' sin BC : R : : sin CD : sin B, (19.) ; and in the triangle
; 4vvi Sin . AC : R : : sin CD : sin A 4 wherefore, by equality in-
; verseJy, sin BC : sin AC : : sin A : sin B. In the same manner) it

may be jHroyqd, that sia BC .; sin AB ; ; sio A ; fiin C, d&c/ TfetW-

fore, &c. Q. £. D.

SPHEatlCAL TRIGONOMETRY.

PROP. XXV.

In obUque angled spherical triangles, a perpendicular arch being drawn
Jram aai^irfthe aflgles upon the opposite side^ the cosines of ike auffhfs .
id the base are proportional to ike sines of die segments (fihe vertical
angle.

Let ABC be a triangle, and the arch CD perpendicular to tbe^ bas^
BA ; the cosine of the angle B will bo to the cosine of the angle A/
as the sine of the angle BCD to the sine of the angle ACD.

For having drawn CD perpendicular to AB, in the right angled
triangle BCD, (23.) cos CD : R : : cos B : sin DOB ; and in the ri^ht
angled triangle ACD, cos CD : R : : cos A r sin ACD ; therefore.
(11. 5.) cos B : sin DCB : : eos A : sin ACD, and alternately, cosB
: coa A : : sin BCD : sin ACD. Q. £. D.

PROP. XXVI.

The smne things remaining, the cosines of the sides BC, CA, are jpro-
portional to the cosines of BD, DA, the s^ments of the base.

For in tbe triangle BCD, C^.), cos BC : cos BD : : cos DC : R,
and in the triangle ACD, cos AC : cos AD : : cos DC : R ; therefore-
(11« 5.) cos BCT? cos BD : : cos AC : cos AD^ and alternately, cos
BC : cos AC : : cos BD : cos AD. Q. £. D.

PROP. xxvn.

The smne. construction remaining^ t?ie sines of BD, DA the segments of
the base are redproeatty proportional to the tangents of B and A, -
^ angles ai the base.

In the triangle BCD, (18.), sin BD : R : : tan DC : tan B ; and in
tbe> triangle AcD, sin AD : R : : tan DC : tan A ; therefore, by equal*
ity inversely, sin BD : sin AD : : tan A : tan B; Qi E. D.

i«

384 SPHERICAL TRIGONOHETRT.

PROP. XXVIII.

ne«ameo(MUtructiiMremaming,theeotine*qfthetegmenttofthever'
tieti angle are redproeaUy proportional to the tangents of the eidet. '

•

Because (21.), cos BCD : R : : tan CD : tan BC, and also cos
ACD : R : : tan CD : tan AC, by equality inversely, cos BCD : cos
ACD ! : tan AC : tan BC. Q. £. D.

PROP. XXIX;

If from an angles of a spherical triangle there he draacn a perpendieu"
lor to the opposite side^or base^ the rectangle cantained by the tan"
gents of half the «*»i, and of half the difference of the segments of
the base is equal to the rectangles contained by the tangents of half the
sum^ and of half the difference of the two sides of the triangle.

Let ABC be a spherical triangle, and let^he arch CD be drawn
from the angle C at right angles to the base AB, tan i {m-jrn) x tan ^
(to— w)=i tan {a+b) X^tan (a—b).

Let BCs=a, AC=^; DD=m, AD=n. Because (26.) cosa : cos
h : : cos m : cos n, (E. 5.) cos a+d : cos a — cos b z : cosm4-cos n :
cos m — cos n. But (1. Cor. 3. PI. Trig.), cos a+cos b : t^os oc-^os b
: : cot J (a+b) : tan -^ (a-^zb), and also, cos m+cos n ; cos m — cos n
: -: cot i (m+») : tan i (m — n). Therefore, (11. 6.) cot f {a+b) :
tan i (a**^) : : cot^ (m+n) : tan j- («— n). And because rectan-
gles of the same ahitude are as their bases, tnn ^ (a+b) X cot j-
{a+b)i tan j- {a+b) X Ian ^ («— ^) : : tan i {m+n) X cot J- (m+n)
: tan i (i»xn)+tan f {m — n)« l^t>w the first and third terms of this
proportion are equal, being each equal to the. square of th^ radius,
(1. Cor. PI. Trig.), therefore the remaining two are equal, (9, 5.) or
Un i {m+n) X, tan ^ (m—n>= tjtn i (a+b) >< tan ^ (a—* ;) that is,
tail 4 (BD+AD) X tan i (BD— AD) = tan i (BC+AC) X tan f
(BC— AC). . Q. E. D. :

Cor. 1. Because the sides of equal rectangles are reciprocally
proportional, tan ^ (BD+AD) : tan ^ (BC+AC) : : tan j (BC—
AC) : tan j (BD— AD).

Colt. 2. Siace, when the perpendicular CD falls within the trian-
gle, BD+AD=3AB, the base ; and wheii CD falls without the trian-
gle BD — ADs=AB, therefore, in the first case, the proportion in the
last corollary becomes tan j (AB) : tan f (BC + AC) : : tan J
(BC-^AC) : tan ^ (BD'-^AD) ; and in the second case, it becomes by
inversion and alternation, tan f (AB) : tan f (BC+AC) : : tao ^
(BC~AC) : tan j- (BD+AD).

SPHERICAL TMGONOMETRT.

2B&

SCHOLIUM-

- • - I

The preceding proposition, whiijh is very tiseful in^pberical trig^,
tiometry, may l>e easily remembered from its analogy to ttre proposi^
tion in plane trigonometry, that the rectangle under half the sum^
and half the deference of the sides of a plane triangle, is equal to the
rectangle under half the sum, and half the difference of the segments
*of the base. See (K. 6.), also 4th Case PI. Tr. We, are indebted
to Napier for this and the two following theorems, which are so well
^apted to calculation by Logarithms, that they must be considered
^as three of the most valuable propositions in Trigonometry.

PROP. XXXi

If a perpencRcidar he draismjrom an angle of a spherical triangle h
the opposite side or base, the sine of the sum if the angles eU the hose
is t9 j&e sine offheit d^erence\as. the tangent of hdlfihe ^aseto the
^UmgeiU of ha^ the difference tf its segments^ when the perpendieuloth
^dlls within ; but as the co-tangent of half the hose to the co-tangetit
'of half the sum of the segments, when the perpendicular fdUs withot^
the triangle .^ And the sine of the sum of the two sides is to the sine
xf their difference as the co4angent of half the angle conttiined by tJte
isideSf to the tangent of hcdf the differenee^f the angles which the per^
pendicttlar makes with the same sides, when it fcUs vnthvn, or to the
tangent of hcSLf the sum of these angles, when it fails without the tri^
Single,

If ABC be^spherical triangle, and AT) a perpendicular to the base
BC, sin (C+B) : sin (0-B) : : tan i BC : tan i (BD-DC,) when
AD falls within the triangle ; but sin (C+B) : sin (C — B) : : cot ^
8C : cot i (BD+DC), when AC faHs without. A«d a^rain,

206

SPHERICAL TRIGONOMETRT.

sin (AB+AC) : sin (AB-AC) : : coU BAG : tan i (BAD -CAD J,
when A D falls within : but when AD falls without the triangle.

sin (AB+AC) : sin (AB-AC) : : cotf BAC : tan j (BAD+CAD).
For in the triangle BAC (27.), tan B : tan C : : sin CD : sin BD,
and therefore (E. 6.), tan C+tajn B i tan C— tan B : : sin BD+sitt
CD : sin BD— sin CD. Now, (by the annexed Lemma) tan C+tan
B : tan C -tan B : : sin (C+B) : sin(C-B), and sin BD+sin CD :
sin BD-sinCD : : tan i (BD+CD) : tan j (BD— CD), (3. PI. Trig.),
therefore, because ratios which are equal to the same ratio are equal
to one another (U. 5.), sin (C+B) : sin (C-B) : : tan J (BD+^D) r
tan i (BD— CD).

Now when Al) is within the triangle, BD+CD=BC, and tfaerefbrer
sin (C+B) : sin (C - B) : : tan i BC : tan i (BD' - CD). And again,
when AD is without the triangle, BD— CD=BC, and therefore sin
(C+B) : sin (C— B) : : tan | (BD+CD) : tan i BC, or because the
tangents of any two arches are reciprocally as their co-tangents^ in
(C+B) : sin (C-.B) : : cot i BC : cot i (BD+CD).

The second part of the proposition is next to be demonstrated. Be-
cause (28.) tan AB : tan AC : : cos CAD : cos BAD, tan AB+tan
AO : tan AB— tan AC : : cos CAD+cos BAD : cos CAD -cos BAD.
But (Lemma) tan AB+tan AC : tan AB-tan AC : : sin (AB+AO) :
sin (AB-AC), and (L cor. 3. PI. Trig.) cos CAD+cos BAD : cos
CAD— cosBAD :: coli(BAD+CAD) : tan i (BAD— CAD). There-
fore (n..6.) sin (AB+AC) : sin (AB— AC) : : cot J (BAD+CAD) :
tan i (BAD— CAD). Now, when AD is within the triangle, BAD+
CAD=BAC, and therefore sin (AB+AC^ : sin (AB— AC) : ; cot i
BAC : tan J (BAD-CAD).

SPHERICAL TRIGONOMETRY. 257

But if AD be without the triangle, BAD— CAD^^BAC, and there-
• fore sin (AB+AC) : sin (AB-AC) : :

tot i (BAD+CAD) : tao i BAG ; or because
cot i (BAD+CAD) : tan i BAG : : cot i BAC :
Un f (BAD+CAD), sin (AB+AG) : sin (AB-AC) : : cot i BAC :
tan + (BAD+CAD). Wherefore, &c* Q, £• D.

LEMMA.

The sum of the tangefOs of emy itoo archesy is to the difference of their
tangents, as the sine of the sum of the archeSy to the sine of their dtf*
ference. ^

Let A and B be two arches, tan A+tan B : tan A-->tan B : : sin
<A+B) : (A— B).

For, by § 6. page 243, sin AXcos B+cos AXsin B=sin (A+B).

and therefore dividing all by cos A cos B, rH «

cos A cos B

sin (A + B) . .' sin A ^ . ^ . . „

s= ~ 4^, that IS, because r = tan A, tan A+tan B

cosAXcosB cosA

sin (A+B) -

= v-— — ~ . In the same manner it is proved that tan A — tan .B

cos AXcos B - ^

= \ L., Therefore tan A + tan B : tan A— tan B : : sin

cos A X cos B

(A+B) : sin (A— B). Q. E. D.

T PROP. XXXL

The sine of half the sum of any two angles of a spherical triangle is to
the sine of half their difference, as the tangent of half the side adjacent
to these angles is to the tangent ofhalfuie difference of the sides op-
posite to them ; and the cosine of half the sum of the same angles is
to the cosine of half their difference, as the tangent of half the side
adjacent to them, to the tangent ofhdfthe sum of the sides opposite.

Xet C+B=aSi G— B=2D, the base BC=2B, and the difference
of the segments of the base, or BD — CD=2X. Thenj because (30.)
sin {C+B) : sin (G— B) : : tan i BC : tan i (BD— GD), sin 2S : sin
2D : : tan B : tan X. Now, sin 2S»sin (S+S)— 2 sin SXcos S,
(Sect. III. cor. PI. Ti.). In the same manner, sin 2D=2 sin DX
cos D. Therefore sin S X cos B : sin D X cos D : : tan B : tan X.

M»

•SPHERICAL TRIGONOMETRY,

Again, in the spberical triangle ABG it haffbeen provec^j that siv
G+sin B : sin C— sii^ B : : sin AB+sin AC : sin AB — sin AC, ancF

«n D : : /sin AB+sin AC : sin AB— sin AC. But (3. PI. Tr.) sin
AB+sin AC : sin AB— sin AC :;tan £ (AB+AC) :tan J (AB-AC) :-
.tan 2 : tan z\, 2 being equal to i (AB+AC) and A to j^ (AB— AC),
Therefore fiin S X cos D : cos S Xsin D : : tan 2 r ton A. Since then
tan X sinDXcosD , tan A cos S Xsin D - «...

, ^ , tanX^tanA (sin D)'»Xcos SXcos D, (sin D)*

quals by equals, ^ — ^X; — ^"=7-^ — STir; stj-^ — f^= / - axV

^ "^ ^ tan B tan 2 (sin S)*Xcos SXcos D (sm S)*

But (29.) ^.ggji^IlggJ^^"'' i (^f +^^-Utis,;^=^i^.

^ ' tan ^ rAR— AC^ tan ^ BG tan A tanB^

and therefore.

tan i (AB— AC)

tan X tan 2 X tan A

> as also

tan A tanB'
tan X tan A tan A*

tan B (tan B)^ * ^ "'"" tan B Ian 2 (tan B)*"
tanX tanA^ (sinD)' ^^^^^(f^^J^^J. ^^^tanA

tan B tan 2 Tsin SY* ' Ttan B'^^ Tsin 8^^ '

tanB
sin D

tan 2 (sin S)'

(tan Bf (sin S)'

tanB

Mill % g _

=-^ — a , or sin S r sin I> : : tan B : tan A, that is,, sin CC+B) ; sin

SID S » > X I /

(C— B) : ; tan ^ BC : tan a (AB— AC) ; >fhicb is the first part of the
proposition. ^

. . . tan A cosSXsinD / tan2 sinSXcosD

_ tan 2 sin SXcos I> ^ tan A cosSXsin D'

- ,, tanX sinDXcosD , « ^ tanX

and since — 5:;=-r-
tan B sm

tan 2 _ (co9 D )^ ^
tan A (cos S)" *

But it was already shewn that :r3-o== — mrr^^<i — > wherefore also

YT^ a- 1 therefore by multiplicatioii* - — s^

DXcosS "^ ^ ^tanB.

tanX tan2_(tan 2)'»

tanB (tan By"

Ian B tan A (tan B)^'

„ tanX^tan 2 (cosDV* , . ^

^^^r ♦~r^^: — 1~7 HT2> as haajust been shewn.

tan B tan A (cos S) •*

-,, - (cosD)^ (tan 2)^ , , cosD

tan 2

,orco»

SPHERICAL TRIGONOMETRY. 260

S : cos D : : tan B : tan 2, that is, cos (C+B) : cos (C— B) : : tan i
BC : tan i (C+B) ; whick is the second part of the proposition.
Therefore, &c. Q. E. D,

Cos. 1. By applying this proposition to the triai^le supplemental
to ABC (11. 3.) and by considering, that the sine of half the sum or
half the difference of the supplements of two arches, is the same with
the sine of half the sum or half the difference of the arches them-
selves : and that the same is true of the cosines, and of the tangents
of half the sum or half the difference of the supplements of two
arches : but that the tangent of half the supplement of an arch is the
same with the cotangent of half the arch itself ^ it will Ibllow, that the
sine of half the sum of any two sides of a spherical triangle, is- to the
sine of half their difference as the cotangent of half the angle con-
tained between them, to the tangent of half the difference ofthe an-
gles oip^osite to them : and also that the cosine of half the sumi of
these sides, is to the cosine of half their difference, ais the cotangent
of half the angle contained between them, to the tangent of half the
sum of the angles opposite to them.

Cos. 2. If therefore A, B, C, be the three angles of a spherical

triangle, a, 5, c the sides opposite to them,

• • • . -^

I. sin i (A+B) f sin i (A— B) : : taiilic : taA i (b— ft).
II. cos i (A+B) : cos i (A — B) : : tan ^ « : tan i (a+ft).
IlL sin i {a+b) : sin i (a— ft) : : tan ^ G : tan ^ (A— B).
IV. cos i (4+6) : cos ^ (a- ft) : : tan i C : tan J <A+B).

870

SPHERICAL TRIGONOMETRT.

PROBLEM I.

In a right angled spheriad triangle, of the three sides and three angles,
any two being given, besides the right angle, to find the other three.

.This problem has sixteen caaes, the solutions of which are con-
tained in the following table, where ABC is any spherical triangle
right angled at A.

OIYSN.

sonofiT.

SOLUTION.

1
2
3

BC and B.

AC.

AB.

R : sin BC : : sin B : sin AC, (19).
R : cos B : : tan BC : tan-AB, (21).
R : cos BC : : tan B : cot C, (20).

AC and C.

AB.

BC.

R : sin AC : : tan C : tan AB, (18).
cos C : R : : tan AC : tan BC, (21).
R : cos AC : : sin C : cos B, (23).

4
5
6

7
8
9

AC and B.

AB.

BC.

tan B : tan AC : : R : sin AB, (18).
sin B : sin AC : : R : sin BC, (19).
cos AC : cos B : : R : sin C, (23).

AC and BC.

AB.
B.
C.

cos AC : cos BC : : R : cos AB, (22).
stn BC : sin AC : : R : sin B, (19).
tan BC:tan AC::R:cbsC, (21).

11
12

13
14
14

15
15
16

AB and AC.

BC.
B.
C.

R : cos AB : : cos AC : cos BC, (22).
sin AB : R : : tan AC : tan B, (18).
sinAC:R::tan AB:tanC, (18).

BandC.

AB.
AC.
BC.

sin B : cos C : : R : cos AB, (23).
sin C : cos B : : R : cos AC, (23).
tan B : cot C : : R : cos BC, (20).

SPHERICAL TRIGONOMETRY.

271

li^ABLE for determining the affections of the Sides and Angles fqund^

by the preceding rules. >

AC and B of the same affection.'

If BCZ90O, AB and B of the same affection, otherwise
different, (Cor. 15.)

If BCZOO^ G and B of the same affection, otherwise dif-
ferent, (15.)

AB and C are of the same affection. (I"^*)

If AC and C are of the same affection, BCZ90<^ ; other-
wise BCZ 90°, (Cor. 16.)
B and AC are of the same affection, (14.)

Ambiguous.
Ambiguous.
Ambiguous.

When BC ^90°, AB and AC of the same ; otherwise of
different affection, (15.)

AC and B of the same affection, (I"^*)

When BCZ90°, AC and C otihe same ; otherwise of dif-
ferent affection, (Cor. 15.)

BCZ90<^, when AB and AC are of the same affection,

(1. Cor. 15.)
'B and AC of the same affection, (14.)

C and AB of the same affection, . (1^0

1
2
3

5
6

7
8
9

10
11

13
14
14

AB and C of the same affection, (14.)

AC and B of the same affection, (I^O

When B and C are of the same affection, BCZ.90^, other-
wise, BC790°i (15.)

15
15

The cases marked ambiguous are those in which the thing sought
has two values, and may either be equal to a certain angle, or to the
supplement of that angle. Of these there are three, in all of which
the things given are a side, and the angle Opposite to it ; and accord-
ingly, it is easy to shew that two right angled spherical triangles
may always be found that have a side and the angle opposite to it the
same in both, but of which the remaining sides, and the remaining
angle of the one, are the supj^ements of the remaining sides and the
remaining angle of the other, each of each.

Though the affection of the arch or angle found may in aU the other
cases be determined by the rules in the second of the preceding tables,
it is of use to remark, that pdl these rules except two, may be reduc

mS SPHESIGAL TRIGONOMSntT.

ed to one»viz. that when the thing found hy the rtdee in the fr^ table U
either a UmgiaU er meoeme; and wkeut of the tangents or cosines em*
ployed in the compukUion qfit^mie ottly behngs to an obtuse angle^ the
angle required is also obtuse,

T^U0, in the Idth ease, when cos AB is found, if C be an obtuse an.
gle, b^ause of cos C, AB must be obtuse ; and in case IB, if either
![ (Mr C be Qbtusci BC is greater than 90^, but if B and C are either
both acute, or both obtuse, BC is less than 90^.

I U isieyident, that this rule does not apply when that which is found
is the sine of an arch ; and this, besides the three ambiguous cases,
bappens alao in other two, viz. the 1st and 11th.- The ambiguity is
obviated, in these two cases, by this rule, that the sides of a spherical
right angled triangle are of 4he same affection with the opposite an*
tfes.

• Two rules are therefore sufficient to remove the ambiguity in all
the cases of the right angled triangle, in which it can possibly be re-
niovedb

SMfiftlCAL TRIGONOMETRY-

6«

tt may be useful to express the same solutions as it the annexed
table, liet A b« at the right angfe^as in the figure, and let the side
opposite to it be a ; let h be the side opposite to B, and^ the side op*
posite to C*

GIVEN.

a and B.

h and C.

b and B.

a andi.

SOUGHT.

h and

B and C.

b.
c.
C.

c.
a.
C.

«.

a*
B.

SOLUTION.

sin A ==s sin « X sin B.
tan c = tan a X i^os B.
cot C = cos a X tan B.

tan t =
tan « =
coeB =

sill 5 X tan C^
tan 6

cos C*

cos b X sin C

sine ^
sin « ==
sinC

tan b

tan 3'
sin A

sin B*
t^osB

cos A *

ffcll

sin c '
sin B
cosC

cos d *

cos b'
sin A

sin d*
tan 6

tan tf*

1 I n

cos d
tanB

tan C =

: cos A X cos f^.
-tan b

sin c*
tanc

sin 6*

cose :
cos b '
cos a =

cos C

sin B*
cosB ~

sin C
cot C

tan B'

■lA .1.

5
6

Id
H

IS
15
16

274

SPHERICAL TRIGONOMUTRIf.

PROBLEM IL

In imp oblique angled spherical triangle^ of the three tides and ^ire&
angles, any three being gtven^ it is required tojind the other three.

In this Table the references («« 4.)9 (c. 5.), di^c. are to (he cases
in the preceding Table, (16.), (27.), dec. to the propositions in Sphe-'
lical Trigonometry.

dlTKM.

Two sides
AB, AC,

and. the in-
cluded angle
A.

SOUGHT.

SOLUTION.

One of the
other angles

The third
side
BC.

Let fall the perpendicular CD from
the unknown angle, not requir-
ed, on AB.

R : cos A : : tan AC : tan AD,
(c. 2.) ; therefore BD is known,
and sin BD : sin AD : : tan A :
tan B, (27.) ; B and A are of
Ihe same or different affection,
according as AB is greater or
less than BD, (16.).

Let fall the perpendicular Ci) from
one of the unknown angles on
the side AB.

R : cos A : : tan AC : Jan AD,
(c. 2.) ; therefore BD is known,
and cos AD : cos BD : : cos
AC : cos BC, (26.) ; according
as the segments AD and DB
are of the same or different af-
fection, AC and CB will be
of the same or different afiec

I tion.

SPHERICAL TRIGONOMETRY.

276

TABLE continued.

OIVSN.

«•«■

SOUGHT.

SOLUTION.

Two angles,
AandACfi,

and

AC,
the side be-
tween them.

The side
BG.

The third

angle

From C the extremity of AC near
the side sought, let fall the per-
pendicular CD on AB.

R ; cos AC : : tan A : cot ACD,
(c. 3.) ; therefore BCD is known,
and cos BCD : cos ACD : : tan
AC : tan EC, (28.)- BC is less
or greater than 90^, according
as the angles A and BCD are
of the same, or difierent affec-
tion.

Let fall the perpendicular CD from
one of the given angles on the
opposite side AB.

R : cos AC : : tan A : cot ACD|
(c. 3.) ; therefore the angle BCD
is given, and sin ACD : sin BCD
: : cos A : cos B, (25.) ; B and
A are of the same or differ-
ent affection, according as CD
fallis within or without the tri-
angle, that is, according as ACB
is greater or less than BCD,
(16.). .

s?^

SPHEwrAT* TniaoKQUSTrvrr^

TABLE continued.

ozysN.

Two sides
AC and BC

The angle

opposite to

Che other gi.

ven side

AC.

and an angle

oppoishe to

one of them,

BC.

SOOOHT.

The angle

ACB

containedbv

the given

sides
AC and BC.

The third
side
AB.

SOLUTION.

Sin BC : sin AC : : »in A : sin B,
(24.) The affection of B is am-
biguous, unless it can be deter,
mined by this rule, that accord-
ing as AC + BC is greater or
less than 180^, A + B is great-
er or less than I80°»(10.)

From ACB the angle sought draw
CD perpendicular to AB ; then
R : cos AC : : tan A : cot ACD,

' (c. 3.) ; arid tan BC : tan AC : :
cos ACD : cos BCD, (28.) ACD
dr BCD = ACB,. and ACB ia
ambiguous, because of the am-
biguous sign + Of — .

Let fall the perpendicular CD from
the angle C, contained by tlie
given sides, upon ther side AB.
R : cos A : : tan AC : tan AD,
(c. 2.) ; cos AC : cos BC : : cos
AD : cos BD, (26.)
AB=AD±BD, wherefore AB
is ambiguous.

SPHERICAL TRIGONOMETRY.

277

TABLE continued.

GIVEN.

Two angles

A,B,
and a side

opposite to

one of them,

B-

SOUGHT.

The three
sides,

AB, AC,
and
BC.

The side
BC

opposite
to the
other

given an*-
gle A.

The side
AB

adjacent
to the
given
angles
A, B.

The third
angle
ACB.

One of the

angles

SOLUTION.

Sin B : sin A : : sin AC : sin BC,
(24.) ; the affection of BC is on-
certain, except when it can be de-
termined by this rule, that accord-
ing as A+B is greater or less than
180°, AC+BC is also greater or
less than 180°, (10.).

From the unknown angle C, draw
CD perpendicular to AB; then
R : cos A : : tan AC : tan AD,
(c. 2.) ; tan B : tan A : : sin AD :
siii BD. BD is ambiguous ; and
therefore AB = AD ± ^P ™ay
have fbur values^ liome of which
will be excluded by this condition
that AB must be less than 180°*

From thfe angle required, C,,draw
CD perpendicular to AB«
R : cos AC : : tan A : cot ACD,
(c. 3.), cos A : cos B : : sin ACD :
sin BCD, (25.). The affection
of BCD is uncertain, and there-
fore ACB = ACD st BCD, has
four values, some of which may
be excluded by the condition, that
ACB is less than 180°.

From C one of the angles not re-
quired, draw CD perpendicular to
AB. Find an arch £ such that
tan i AB : tan i (AC+BC) -. : tan
J (AC— BC) : tan i E ; then, if
AB be greater than E, AB is the
sum, and £ the difference of AD
and DB ; but if AB be less than
E, £ is the sum and AB the difie»
rence of AD, DB, (29.). In ei-
ther case, AlJand BD are known,
and tan AC : tan AD : : R : cos A.

278

SPHERICAL TRIGONOMETRY.

TABLE continued.

OITEN.

The three

angles
A, B, C.

SOUGHT.

One of the
• sides
PC. ',

SOLUTION.

Suppose the supplements of the
three given angles, A, B, C, to
be a, hi c, and to Mb the sides
6f a spherical triangle. Find,
by the 4ast case, the angle of
this triangle, opposite to the
side a, and it will be the sup-
plement of the aide of the given
triangle opposite, to the angle
A, that is, of BC, (IL); and
therefore BC is found.

In the foregoing table, the rules are given for ascertaining the ajf^
fection of the arch or angle found, whenever it can be done : Most
of these rules are contained in this one rule, which ia of general ap#
plication, viz. that when the thing f^nd %$ either a tangent or a cosine,
and of the tangents or cosines empioyed in tht compulation <^itf eithet
one or three belong to obtuse angles, the angle found is also obtuse%
This rule is pairticularly to be attended to in cases 5*and 7, where i|
removes part of the ambiguity.

It may be necessary to remark with respect to the 11th case, that
the segments of the base computed there are those Cut offby the near^^
est perpendicular ; and also, that when the sum of the sides is lesa
|han 180°, the least segment is adjacent to the least side of th.e tri$iu-v
gle : otherwise to the greatest, (17.}«

SPHERICAL TRIGONOMET HT^

27d.

The last table may also be conyeniently expressed ia the following
taanner, denoting thie side oppoisite to the angle A, by a, to B by J,
and to C by c ; and also the segments of the base, or of opposite an-
gle, by X anB y,

GIVEN.

Two sides

h and Cy and

the angle

between

them A.

Angles
AandC

and
side b.

Sides
a and h

and
angle A.

SOUGHT.

SOLUTION.

Find 07, so that

tan d?=tan ftXcos A ; then

. D sin^Xtan A

tan is=:-r-; — ^— — .

sm (c — X) '

then cos a

Find «, ad above,
^ cosft Xcos (c — x)

cos X.

•

Find Xy so that
cot a?=cos ( Xtan A : then

tan^Xcosa;

tan a= -.

cos (c — ^a?)

Find Xy as above,

*u^ n ^^ AXsin (c—x)
then COST B= r— — ^^ -.

sm X

sin B=

sin ftXsid A

sin a

Find a?, so that ,

cot a;^cos h Xtan A : then

^ cos a; Xtan ft

cos C=? ^ -*

tan a

Find Xy so that

tan a;=stan ft X cos A ; and find

1^, so that

CosaXcos X
cos y=-

cos ft

<5=a?± y.

880

SME&ICAL TUiGONOMEf Rt-

TABLE contiDued.

GIVE*.

The angles

AandB

and the

side h.

dy hy '€•

A, B, C.

dOUOHT.

SOLUTION.

sin^XsinA
sin !<=-.-*— — :=.

sin B

Find Xy so that

tan x=tan 6Xcos A ; and y^ so

that

sin^tXtanA
sm y=s-

tan B

c=3:dzy.

Find ap, sorthat

cot a; =008 b Xtan A ; and also 5^,

so Uiat

sin a;Xcos B
sin y=-

c=as.^y.

cos A

Let a+b+c=i8.

1 * ^V'sin (jg — fe)X sin ('i* — c)

y<sin 6 X sin e

or cos

, V' sin i* 9 X sin (^»— a)
i-A= .^

^sin 6Xsin c

Let A+B4-C=cS.

Q. v^cos i SXcos a S— A)
Sm ■J' c^s ■ ■ ..

^sin B XainO

or cos

. V cos(+S— B)icps(S— C)

ia^ — : — ====5===^^

^sin BXsin O

^k^mmm^mm—^mtm-^

APPENDIX

SPHERICAL

TRIGONOMETRY,

CONTAINING ^

NAPIER'S RULES OF THE CIRCULAR PARTS.

The rule of the Circular Parts, invented by Napieb, is of great
use in Spherical Trigonometry, by reducing all the theorems em-
ployed in the solution of right angled triangles to two. These two
are not new propositions, but are merely enunciations, which, by help
of a particular arrangement and classification of the parts of a trian^e,
include all the six propositions, with their corollaries, which have been
demonstrated above from the 18th to the 23d inclusive. They are
perhaps the happiest example of artificial memory that ia known*

DEFINITIONS.
L

If in a apherical triangle, we set aside the ridit ang^, and considef
only the Bwe remaining parts of the triaogie, viz* the three sides
and the two oblii|iie aii^esi» then the two sides which contain the
right an^e, and the complements of the other three, namelv, of the
two angles and the hypotenuse, are called the Cireidar Pari$0

Thus, in the triangle ABC right angled at A, the circular parts are
AC, AB with the complements of B, EC, and C« These parts are
ealled circular ; because, when they are named in the aatural order
of their successioo, they go round the triangle.

11.

Whenofthefive circular parts any one is taken, for the middle part^
theo of the remaining fiwr, the two which are imnnediately adjaeenC
to it, on tbe right and left, are called the adjacent parts ; and the
other two, eadb of which is separated from the middle by au a4ia>
cent part, aue called opposite parts*

5BB2 APPENDIX TO

Thus in the right angled triangle ABC, A> being the right angle, AC
AB, 90<»--B, 90^— BC, 90°— C, are the circular parts, by Def. 1. ;
and if any one, as AC, be reckoned the middle part, then AB and 90^^
— C, whictrare contiguous to it on different sides, are called adjacent
parts ; and 90'' — B, ^^ -BC are the opposite parts. In like manner

if AB iar taken for the middle part, AC and 90<>— B are the adjacent
parts : 90°- BC, and 90°— C are the opposite. Or if 90O--BC be the
middle part, 90° — B, 90° — C are adjacent ^ AC and AB opposite, &e«
This arrangement being made, the rule of the circular part is con-
tained in the following

PROPOSITION.

In a right 'angled spherical triangle, the rectangle under the radius and
the sine of the middle part, is equal to the rectangle under the tangents
qf the adjacent parts ; w to the rectangle under the cosines ^ the

I opposite parts.

The truth of the two theorems included in this enunciation may be
easily proved, by taking each of the five circular parts in succession
for the middle part, when the general proposition will be found to co-
incide with some one of the analogies in the table already given for
the resolution of the cases of right angled spherical triangles. Thus,^
in the triangle ABC, if the complement of the hypotenuse BC be tak*
en as the middle part, 90° — B, and 90° — C, are the adjacent parts,
AB and AC the opposite. Then the general rule gives these two
theorems, RXcos BC=cot BXcot C ; and RXcos BCs^cos ABX
cos AC. The former of these coincides. with the cor. to the 20th ;
and the latter with the 22d.

To apply the foregoing general proposition, to resolve any case of
a right angled spherical triangle, consider which of the three quanti-
ties named (the two things given and the one required) must be
made the middle term, in order that the other two may be equi-
distant from it, that is^ may be both adjacent, or both opposite ; then
one or other of the two theorems contained in the above enunciation^
will give the value of the thing required.

Suppose, for example, that AB and BC are given, to find C ; it is
evident that if AB be nade the middle part, BC and C are the oppet^

SPHERICAL TRIGONOMETRY. 28S

site parts, and therefore RXsin AB=sin CXsin BC, for sin C=cos

(90O— C), and cos (90o— BC) = sin BC, and consequently

. ^ sin AB
am 0=5-^ — ^/7.
• sm D\j

Again, suppose that BC and C are given to find AC ; it is obvious

that C is in the middle between the adjacent parts AC and (90^— BC),

therefore R Xcos C=tan AC Xcot BC, or tan AC= Tr;^=cos C+

cotBC

tan BC ; because, as hai^ been shewn above, -wr:7-=tan BC.

cot BC

In the same way may all the other cases be resolved. One or two
trials will always lead to the knowledge of the part which in any given
case is to be assumed as the middle partj and a little practice ;wili
make it easy, even without such trials, to judge at once which of them
is to be so assumed. It may be useful for the learner to range the
names of the five circular parts of the triangle round the circumfer-
ence of a circle, at equal distances from one another, by which means
the middle part will be immediately determined.

Besides the rule of the circular parts, Napier derived from the last
of the three theorems ascribed to him above, (schol. 29.) the solu-
tions of all the eases of oblique angled triangles. These solutions are
as follows : A, B, C, denoting the three triangles of a spherical trian-
gle, and a, 5, c, the sides opposite to them.

Given two sides 5, c, and the angle A between them.
To find the angles B and C*

tan i (B— C)=cot i A X ^l^ii^Zli). (31.) cor. 1.

sm t (o+c) *

tanHB+C)=cot-AX^i-l^. (31.) cor. 1.

To find the third side a, i>

sin B : sin A : : sin 6 : sin a.

II.

Given the two sides &, c, and the angle B opposite to one of them.

To find C, and the angle opposite to the other side.

sin 6 : sin c : : sin B : sin C.

To find the contained angle A.

eo. * A=:t«» + (B-C)x!|l|-Jg). (31.) cor. 1.

284 APPENDIX TO

To find the third side a.
^ Sin B : sin A : : sin 6 : sib a.

III.

Given two angles A and B, and the side e between them.

To find the other two sides a, h.

tan + (J_«)=tanieX-^i|^^. (31.)
tani(»+«)=taBieX?2jii^j. (3l'.)

To find the third angle C.
sin a : din c : : sin A : sin C.

Given two angles A and B, and the side a, opposite to one of them.

To find by the side opposite to the other,
sin A : sin B : : sin a : sin h.

To find Cy the side between the given angles*

».J.-i(.-*)xli|-{^,. (31.)

To find the third angle C.

sin a : sin c : : sin A, : sin C«

The other two cases, when the three sides are given to find the an-
gles, or when the three angles are given to find the sides, are resolved
by the 29th, (the first of Napier's Propositions,) in the same way as
in the table already given for the case of the oblique angled triangle.

There is a solution of the case of the three sides being given, which
it is often very convenient to use, and which is set down here, though
the proposition on which it depends has not beien dempAStrated.

SPHERICAL TRIGONOMETRY-

S85

Let a, b, c, be the three given sides, to find the angle A, contained
between b and c.

If Rad. = 1, and a + 6 + « = «,

^sin 6 X sin c

1 A _ l/sin (i * X sin j (* — a)

v/sin^Xsinc

In like manner, if the three angles, A, B, C are given to find c, the
side between A and B.
LetA + B + C = S,

cos

sin

V^cos ^ S X cos (J S — A) ,

V'stnB X sin C
j^^ ^ v-'cos (^S -^ B) X cos (4S - C)

or,

cos^c =

^sin B X sin C

These theorems, on account of the facility with which Logarithms
are applied to them, are the most convenient of any for resolving the
two cases to which they refer. When A is a very obtuse angle, the
second theorem, which gives the value of the cosine of its half, is to
be used ; otherwise the first theorem, giving the value of the sine of
its half is preferable. The same is to be observed with respect ta
the side c, the reason of which was explained, Plane Trig* SchoL

Enn OF SPHERICAL TSI60N0METRT.

NOTES

ON THB

ELEMENTS.

■MM

rtMMiV^««l*«lta*MV

NOTES

ON Tils

FIRST BOOK OF THE ELEMENTS.

• ' - I ^

DEFINITIONS*
I.

Ln the definitions a few chairged have b^en mad&) of wMeh It i^
Irecessary to gtvi9» some account. One of these changes respects
the first definition, that of a point, which Euclid has said to be, *^ That
which has no patts, of which has no magnitude^' Now, it has beeti
objected to this definition, that it contains only a negativifr, aildthatit is
not convertible, as «3very good definition ought certainly to be^ That
it is not convertiblois evident, foir though every point is utiextended, o^
witHout magnitude, yet every thing unextended or without Magnitude^
is not a point. To this it is impossible to reply, and therefore it be<»
comes necessary to change the definition altogether, which is accord-^
itigly done here, a point being defined to be, tHoA which has position hut
iwt magnitude. Here the affirmative part includes all that is es^
oential to a point, and the negative part includes every thing that id
not essential to it. I am indebted for this definitioh to a friend, by
whose judicious find teamed remarks I have often profited*

II.

Alfter the second definition Euclid ho^ Introduced the following}
«* the extremities of a line ^re points."

Now, this is certainly not a definition, but ah inference troto the de»
finitioQs of a porat and of a line. That which terminated a line cail
have no breadtfi, as the line in which it is has none ; and it can haVe no
length, as it would not then be a termination, but a pan of that which
it is supposed to terminate. The termination of a line Can therefore
have no magnitude, and having necessarily positioti, it is a point. But
as it is plain, that in all this we are drawing a consequehce from two
definitions already laid down, and not giving a neW definition,! have
taken the liberty of putting it down as a corollary to the second defi*
nitioh, and have added, that the intersections of one line with anothef

.37

290

NOTEa

are points, as this affords a good illustration of the nature of a point,.
and is an inference exactly of the same kind ^ith the preceding. The
same thing nearly has been done with the fourth 'definition, where that
which Euclid gave as a separate definition is made a corollary to the
fourth, because it is in fact an inference deduced from comparing the
definitions of a superficies and a line.

As it is impossible to explain the relation of a superficies, a line,,
and a point to one another, and to the solid in which they all originate,
better than Dr. Sirason has done, I shall here add, with very little
change, the illustration given by that excellent Geometer.

'' lit is necessary to consider a solid, that is, a magnitude which haa
length, breadth, and thickness, in order to understand aright the defi*
nitions of a point, line and superficies } for these all arise from a solid,
and exist in it ; The boundary, or boundaries which contain a solid,
are called superficies, or the boundary which is common to twc^ solids
which are contiguous, or which divides one solid into two contiguoui^
parts, is called a superficies; Thus, if BCGFbeoneof theboundal-ies
which contein the solid ABCDEFGH, or which is tfee common hout^
dary of this solid, and tjie solid BKJUCFNMG, and is therefore in the
one as' well as the other solid, it is called a superficies, and has no^
thickness ; For if it have any, this thickness must either be a part of
the thickness of the solid AG, or the solid BM, or a part of the thick,
ness of each of them. It cannotbe a part of the thickness of the solid
BM ; because, if this solid be removed fram the solid AG, the superfi-
cies BCGF, the boundary of the solid AG, reinairis still the same as it
was. Nor can it be a part of the thickness of the s^Ud AG : because
if this be removed from the solfd BM, the superficies BCGF, the boun-
dary of the solid BM, does nevertheless remain ; therefore the* su-
perficies BCGF has no thickness, but only length and breadth.

** The boundary of a superficies is calleda lino; oralineisthecom-
inon boundary of two superficies that are cooti^iwus, or it is that which
divides one superficies into two contiguous parts : Thus, if BC be one»
of the boundaries which contain the superficies ABCD, or which is the
<5ommon boundary of this superficies, and of the superficies KBCL,
which is contiguous to it, this boundary BC is called a line, and has no
>readth ; For, if it have any, this must be part, either of the breadth
of the superficies ABCD or of
the superficies KBCI^, or part of
«ach of them. It is not part of
the breadth of the superficies
KBCL ; for if this superficies
be removed from the superficies
ABCD, the line BC which is the
boundary of the superficies AfeCD
remains the same as it was. Nor
can the breadth that BC is sup-
posed to have, be a part of the
breadth of the superficies ABCD ;
because, if this be removed from,

NOTES.

S91

tfa« superficies KBCL, the line BC, which U the boundary of the su-
perficies KBCL, does nevertheless remain : Therefore the line BC
has no breadth. And because the line BC is in a superficies, and
that a superficies has no thickpess, as was shown ; therefore a line
has neither breiadth nor thickness, but only length.

*^ The boundary of a line is called a point, or a point is a common
boundary or extremity of two lines that are contiguous : Thus, if B
be the extremity of the line AB, or the common extremity of the two
tines AB, KB, this extremity is called a point, and has no length : For
if it have any^ this length must ei-

ther be part of the length of the
line AB, or of the line KB. It is
not part of the length of KB ; for
•f the line KB be removed from
AB, the point B, which is the ex-
tremity of tlie line AB, remains
the same as it was ; Nor is it part
of the length of the line AB ; for
if AB be removed from the line
KB, the point B, Which is the ex-
tremity of the line KB, does ne«
vertheless remain : Therefore the
point B ha^ no length ; And because a point is in a line, and a line
has neither breadth nor thickness, therefore a point has no lengthy
breadth, aor thickness* And in this manner the definition of a point,
line, and superficies are to be understood."

in.

£ueUd has defined a straight line to be a line which (as we translato
it) ^' lies evenly between its extreme points." This definition ig ob*
viously faulty, the word evenly standing as much in need of an ex]|)la^
nation as the word straight, which it is intended to define. In the
original, however, it must be^confessed, that this inaccuracy is at least
less striking than in our translation ; for the word which we render
evenly is s^kTs, equally ^ and is accordingly translated ex (Bquo, and efua-
liter by Commandine and Gregory. The deifinition, therefore, is, that
a straight line is one which lies equally between its extreme points :
and if by this we understand a line that lies between its extreme points
so as to be related exactly alike to the space on the one side of it, and
to the Tspace on the other, We have a definition that is perhaps a little
too metaphysical, but which certainly contains in it the essential cha-
racter of a straight line. That Euclid took the definition in this sense,
however, is not certain, because he has not attempted to deduce from
it any property whatsoever of a straight line ; and indeed, it should
seem not easy to do so, without ^n^ploying some reasonings of a more
metaphysical kind than he has any where admitted into his Elements.

292 NOTES.

T6 supply the defects of .his definitioti, he has therefore introdueed
the Ai[ioin, that tvx) straight Unci cannot inclose a space ; on which
Axiom it is, rdcI not on his definition of a straight line, that his de-
monstrations are fbunded. As this .manner of proceeding is cer.
tainfly not so regular and seientific as that of laying down a deiinition,
from which the properties of the thing defined may be logically de-
duced, I have substituted another definition of a straight line in the
room of Euclid's. This definition of a straight line was suggested by
a remark of Boseovich, who, in his Notes on the philosophical Poetvt
of Professor Stay, says, << Rectam lineam recte congruere totam toti
** in infinitum productum si bina puncta unius binis aheriuscongraant,
^' patet ex ipsa admodum clara rectitudinis idea quam habemns.^
(Supplementum in lib. 3. § 550.) Now, thai which Mr. Boscovick
would consider as an inference from our idea of straightness, seems
itself to be the essence of that idea, and to afford the best criterion
forjudging whether any given line be straight or not. On this prin-
ciple we have given the definition above, ijf there be two lines which
ca^mot coincide in ttco pointSy mlhovt coinciding altogether ^ eaehof them'
is caUed a straight line. .

This definition was otherwise expressed in the two former editions ;
It was said, that h'nes are straight lines which cannot coincide in part,
without coinciding altogether. This was liable to an objection, viz.
. that it defined straight lines, but not a straight Une ;' and though this
in truth is but a mere cavil, it is better to leave no room for it. The
definition in the form now given is also more simple.

From the same definition, the proposition which Euclid gives as an
Axiom^ that two straight lines cannot inclose a space, follows as a
necessary consequence. For, if two lines inclose a space, they must
intersect one another in two points, and yet, in the intermediate parti
must not coincide ; and therefore by thfe definition they are not
straight lines. It follows in the same way, that two straight lines
cannot have a common segment, or cannot coincide in part, without
coinciding ahogether.

"'After laying down the definition of a straight line, as in the first
Edition, I was favoured by Dr. Reid of Glasgow with the perusal of a
MS. containing many excellent observations on the first Book of Eu-
clid, such as might be expected from a philosopher distinguished for
the accuracy as well as the extent of his knowledge. He there de-
fined a straight line nearly as has been done here, viz. « A straight
•* line is that which cannot meet another straight line in more points
♦* th^n one, otherwise they perfectly coincide, and are one and the
*« same." Dr. Reid also contends, that this must have been Euclid's
own definition ; because, in the first proposition of the eleventh Book,
that author argues, "that two straight lines cannot have a common seg-
" ment, for this reason, that a straight liiie does not meet a straight
" line in more points than one, otherwise they coincide.'* Whether
this amounts to a proof of the definition above having been actually
Euclid's^ I will not takoupon me to decide ; but it is certainly a proof

. NOTES. 208

that the i^ritings of that Geometer ought long since to have saggested
this definition to his conmientators ; and it reminds me, that 1 might
have learned from these writings what I have acknowledged ahove
to be derived from a remoter source. ^

There is another characteristic, and obvious property of straight
lines, by which I have often thought that they might be very conve-
niently defined, viz. that the position of the whole of a straight line
is determined by the position of two of its points, in so much that, when
two points of a straight line continue fixed, the line itself cannot change
its position. It might therefore be said, that a straight Une is one in
tphtchf if the position of two points be determined, <Ae position of the
whole line is determined. But this definition, though it amount in fact
to the same thing with that already given, is rather more abstract, and
-not so easily made (he foundation of reasoning. I therefore thought
it best to lay it aside, and to adopt the definition given in the text.

The definition of a plane is given from Dr. Simson, Euclid's b^ing
liable to the same objections with his definition of a straight line ; ifbr,
he says, that a plane superficies is one which ** lies evenly between
'' its extreme lines." The defects of this definition are completely r^
moved in that which Dr. Simson has given. Another definition difiTer-
ent from both might have been adopted, viz. That those superficies
are called plane, which are such, that if three points of the one coin-
cide with.three points of the other, the whole of the one must coin-
cide with the whole of the other. This definition, as it resembles
that of a straight line, already given, might, perhaps, have been in-
troduced with some advantage ; but as the purposes of demonstration
cannot be better answered than by that in the text, it has been thought
best to make no farther alteration.

VI.

In Euclid, the general definition of a plane angle is placed before
that of a rectilineal angle, and is meant to comprehend those angles
which are formed by the meeting of the other lines than straight
lines. A plane angle is said to be " the inclination of two lines to
" one another which meet together, but are pot in the same direc
" tion." This definition is omitted here, because that the angles form-
ed by the meeting of curve lines, though they may become the sub-
ject of geometrical investigation, certainly do not belong to the Ele-
ments ; for the angles that must first be considered are those made
by the intersection of straight lines with one another. The angles
formed by the contact or intersection of a straight line and a circle,
or of two circles, or two curves of any kind with one another, could
produce nothing but perplexity to. beginners, and cannot possibly be
understood till th9 properties of rectilineal angles have beep fully

r2M

NOTES.

explaiiied. On this ground, I am of opinion, that in an elementary
treatise, it may /airly be omitted. Whatever is not useful, should,
in explaining the elements of a science, be kept out of sight ahoge-
ther ; for, if it does jiot assist the progress of the understanding, it
will certainly retard it. .

AXIOMS.

Among the Axioms there have been made only two alterations.
The 10th Axiom in Euclid is, that ^^ two straight lines cannot inclose
'^ a space ;" which, having become a corollary to our definition of a
straight line, ceases of course to be ranked with self-evident propo-
>8itions. It is. therefore removed from among the Axioms, and that
which was before the 11th is accounted the 10th.

The 12th Axiom of Euclid is, that^^' if a straight line meets two
" straight lines, so as to make the two interior angles on the same
'< side of it taken together less than two right angles, these straight
''^ lines- l)eing continually produced, shall at length meet upon that
^< side on which are the angles which are less than two right angles."
Ikistead of this proposition, which, (hough true, is by no means self-
-evident ; another that appeared more obvious, and better entitled to
i>e accounted an Axiom, has been introduced, viz. " that two straight
^Mines, which intersect one another, cannot be both parallel to the*
^' same straight line." On this subject, however,' a fuller explanation
is necessary, for which see the note on the 29th Prop.

PROP. IV. and VIIL B. I.

. The fourth and eighth propositions of the first book are the foun-
dation of all that follows with respect to the comparison of triangles.
They are demonstrated by what is called the method of supraposition,
that is, by laying the one triangle upon the other, and proving that
they must coincide. To this some objections have been made, as if
it were ungeometrical to suppose one figure to be removed from its
place and applied to another figure. " The laying," says Mr. Thomas
Simson in his Elements^ <<of one figure upon another, whatever evi-
^' dence it may afford, is a mecJianical consideration, and depends on
^'no postulate." It is notclear what Mr. Simson meant here by
the word mechanical .^but he probably intended only to say, that the
method of supraposition involves the idea of motion, which belongs
rather to mechanics than geometry ; for I think it is impossible that
such a Geometer as he was could mean to assert, that the evidence
derived from this method is like that which arises from the use of
instruments, and of the same kind with what is furnished by experi-
ence and observation. The demonstrations of the fourth and eighth,
as they are given by Euclid, are as certainly a process of pure rea-
fi(mipg> depending solely on the idea of equality, us established in the

NOTES. 305'

8th Axiom, as aiiy thing in geometry. But, if still the removal of the
triangle from its place be considered as creating a difficulty, and as
inelegant, because it involves an idea, that of motion, not essential to
geometry, this defect may be entirely remedied, provided that, to
Euclid's three postulates, we be allowed to add the following, viz.
That if there he two eqwd straight lines y and if any figure whalsoener
he (^onstUiUed an the one^ a figure every way equal to it may heconstUui'
ed on the other. Thusrf A£ and D£ be two^qual straight lines, and
ABC a triangle on the base AB, a triangle DEF every way equal to
ABC may be supposed to be constituted on DE as a base. By this
it is hot meant to assert that the method of describing the triangle
DEF is actually known, but merely that the triangle DEF may be
conceived to exist in all respects equal to the triangle ABC. Now,
there is no truth whatsoever that is better entitled than this to be rank-
ed among the Postulates or Axioms of geometry ; for the straight
lines AB and DE being every way equal, there can be nothing be-
longing to the one that may not also belong to the other.

On the strength of this Postulate the fourth proposition is thus de-
monstrated.

If ABC, DEF be two triangles, such that the two sides AB and AC
of the one are equal to the two ED, DF of the other, and the angle
BAC, contained by the sides AB, AC of the one, equal to the angle'
EDF, contained by the sides ED, DF of the other ; the triangles ABC
and EDF are every way equal.

4 D

On AB let a triangle be constituted every way equal to the triangle
DEF ; then if this triangle Coincide with the triangle ABC, it is evi*'
dent that the proposition is true, for it is equal to DEF by hypothesis,,
and to ABC, because it coincides with it; wherefore ABC, DEF are
equal to one another. But if it does not coincide with ABC, let it
have the position ABG ; and first suppose G not to fall on AC ; then
the angle BAG is not equal to the angle BAG. But the angle BAG is
equal to the angle EDF, therefore EDF iind ABC are not equal, and
they are also equal bjr^ hypothesis, which is impossible. Therefore
the point G must fall upon AC ; now, if it fall upon AC but not at C„
'then AG is not equal to AC ; but AG is equal to DF, therefore DF
and AC are not equal, and they are also equal by supposition, which
is impossible. Therefore' G must Coincide with C, and the triangle
AGB with the triangle ACB. But AGB is every way equal to DEF,^
therefore ACB and DEF are also every way equal. Q. E. D.

NOTES.

By help of the same postulate, the 5th may also be very easily de«
moDstrated.

.Let ABC be an isosceles triangle, in which AB, AC are the equal
sides ; the angle ABC, ACB opposite to these sides are also equal*

Draw the straight line £F equal to BC, and suppose that on £F
the triangle D£F is constituted every waj equal to the triangle ABCi
that is, having DE equal to AB, DF to AC, the angle EDF to th«
angle BAC, the angle ACB to the angle DFE, &c.

Then because DE is equal to AB, and AB is equal to AC, DE ii
equal to AC ; and foi* the same reason, DF is equal to AB. And be-*
cause DF is equal to AB, DE to AC, and the angle FDE to the an<yle
BAC, the angle ABC is equal to the angle DFE (4. 1.). But The
angle ACB is also, by hypothesis, equal to the angle DFE ; there-
fore the angles ABC, ACB are equal to one another. CI. E. D.

Thus also, the 8th proposition may be demonstrated independently
of the 7th.

Let ABC, DEF be two triangles, of which the sides AB, AC are
equal to the sides DE, DF each to each, and also the base BC to the
base EF ; the angle BAC is equal to the angle EDF.

A P

NOTEa

s&?

On BC, which is equal to EF, and on the side of it opposite to the
IriaDgle ABC, let a triangle BGC be constituted every way eoual to
the triangle DEF, that is, having GB equal to DE, GC to DF, tl^e
angla BGC to the angle EDF, <fec. : join AG.

Because GB and AB are each equal, by hypothesis, to DE, AB and
OB are equal to one -another, and the triangle ABG is isosceles*
Wherefore also (5, 1.) the angle BAG is equal to the angle BOA. In
the same way, it is shown that AC is equal to GC, and the angle CAG
to the angle CO A. Therefore adding equals to equals, the two angles
BAG, CAG together are equal to the two angles BOA, CGA together,
that is, the whole angle BAG to the whole BOX). But the angle BGG
is, by hypothesis, -equal to the angle EDF, therefore also the angle
BAG is equal to the angle EDF. Q. E. D.

Such demonstrations, it must, however, be acknowledged, trespass
against a rule which Euclid has uniformly adhered to throughout the
Elements, e:fccept where he was forced by necessity to depart from it;
This rule is, that nothing is ever supposed to be done, the manner of
doing which has not been already taught, so that the construction is
derived either directly from the three postulates laid down in the be-
ginning, or from problems already reduced to those postulates*^ Now>,
this rule is not essential to geometrical dciiiunBtratioOt where, for the
purpose of discovering the properties of figures, we are certainly at
liberty to suppose any figure to be constructed, or any line to be drawn,
the existence of which does not involve an impossibility. The only
use, thereibre, of Euclid's rule is to guard against the introduction of
impossible hypotheses, or the ticking for granted that a thing may exist
which in fact implies contradiction; from such suppositions, false
conclusions might, no doubt, be deduced, and the rule is therefore
useful in as much as it answers the purpose of excluding them. But
the foregoing postulatum could never lead to suppose the actual exis-
tetice of any .thing that is impossible ; for it only assumes the existence
of a figure equal and similar to on^ already existing, but in a different
part of space from it, oriiaving one of its sides in an assigned position.
As there iiSi no impossibility in the existence of one of these figures, it
is evident that there can be none in the Existence of the other.

PROP. VII.

Dr. Simson has very properly changed the enunciation of this pro*
position, which, as it stands in the original^ is considerably embarrass*
ed and obscure. His enunciation, with very little variation, is re-
tained here.

PROP. XXI.

It is essential to the truth of this proposition, that the straight lines
drawn to the point within the triangl^ be drawn from the two extremi-
ties of the base ; for, if they be drawn from other points of the base,
their sum may exceed the sum of the sides of the triang^in any ratio
that is less than that of two to one. This is demonstrated by Pappus

2M NOTES.

Alezandrinus in the 3d Book of bis Mathematical Collections^ but tb#
demonstration is oFa kind that does not belong to this place. If it bo
required simply to show, that in certain cases the sum of the two lines
drawn to the point within the triangle may exceed the sum of the sides,
of the triangle^ the demonstration is easy, and is given nearly as fol-
lows by Pappus, and also by Proclus, in the. 4th Book of his Com-
mentary on Euclid.

Let ABC be a triangle, having the angle at A a right angle t let D
be any point in AB ; join CD, then CD will be greater than AC, be-
cause in the triangle ACD the angle CAD is greater than the angle
ADC. FromDC cut off DE equal to AC ;
bisect CE in F, and join BF ; BF and
Fl) are greater than BC and CA.

Because CF is equal to FE, CF and
I'B are equal to EF and FB, but CF
and FB are greater than BC, therefore
£F and FB are greater than BC. . To
EF and FB add EB, and to BC add AC, J^
which is equal to ED by construction,
a^d BF and FD will be greater than BC and CA. Q. E. D.

It is evident, that if the angle BAC be obtuse, the same reasoning
may be applied.

This proposition is a sufficient vindication of Euclid for having de-
monstrated the 21st proposition, which some affect to consider as self-
evident ; for it proves that the circumstance on which the truth of
that proposition depends is not obvious, nor that which at first sight
it is supposed to be, viz. that of the one triangle being included within
the other. For this reason I cannot agree with M. Clairaut, that Eu-
did demonstrated this proposition only to avoid the cavils of the So-
phists^ But I must, at the same time, observe, that what the French
Geometer has said on the subject has certainly been misunderstood,
and, in one respect, unjustly censured by Dr. Simson. The exact
translation of his words is as follows : << If Euclid has taken the trou-
*' ble to demonstrate, that a triangle included within another has the
*^ sum of it sides less thai^the sum of the sides of the triangle in
" which it is included, we are not to be surprised. That Geometer
'< had to do with those obstinate Sophists, who made a point of refus-
<< ing their assent to the most evident truths," dec. (Elements de Geo-
metrie par M. Clairaut. Pref.)

Dr. Simson supposes M. Clairaut to mean,^ by the proposition
which he enunciates here, that when one triangle is included in an-
other, the sumofjthe two sides of the included triangle is necessarily
less than the sum of the two sides of Che triangle in which it is
included, whether they be on the same base or not. Now this is not
only not Euclid's proposition, as Dr. Simson remarks, butitis not true,
and is directly contrary to what has just been demonstrated from Pro-
clus. Bat the fact seems^ to be, that M. Clairaut's meaning is entirely
different, and that he intends to speak not of two of the sides* of a tri-
angle, but of all the three ; so that his proposition is^ " that when one

NOTES. 299

^ triangle is included within another, the sum of all the three sides of
*< the included triangle is less than the sum of all the three sides of the
" other/' and this is without doubt true, though I think by no means
self-evident. It must be acknowledged also, diat it is not exactly
Euclid's proposition, which, however, it comprehends under it, and is
the general theorem, of which the other is only a particular case.
Therefore, though M. Clairaiit may be blamed for maintaining that
to be an Axiom which requires demonstration, yet he is not to be
accused of mistaking a false proposition for a true one.

PROP.XXII.

Thomas Simpson in his Elements has objected to Euclid's demon-
stration of this proposition, because it contains no proof, that the two
circles made use of in the construction of the Problem mUst cut one
another ; and Dr. Simson, on the other hand, dways unwilling to ac-
knowledge the smallest blemish in the works of Euclid, contends that
the demonstration is perfect. The truth, however, certainly is, that
the demonstration admits of some improvement ; for the limitation
that is made in the enunciation of any Problem ought always to be
shewn to be necessarily connected with the construction of it, and this
is what Euclid has neglected to do in the present instance. The de-
fect may easily be supplied, and Dr. Simson himself has done it in
reflect in his note on this proposition, though he denies It to be neces-
sary.

Because that of the three straight lines DF, F6, GH, any two are
greater than the third, by hypothesis, FD is less than F6 and GH, that
is, than F^H, and therefore the circle described from the centre F, with
the distance FD must meet the line FE between F and H ; and, for

the like reason, the circle described from the centre G at the distance
GH, must meet DG between D and G, and therefore the one of these
circles cannot be wholly within the other. Neither can the one be
wholly without the other, because DF and GH are greater than FG;
the two circles must therefore intersect one anbther.

am NOTES.

PROP. XXVII. and XXVIII.

Euclid has been guilty of a slight inaccuracy in the enunciations irf
these propositions, by omitting the condition, that the two stratghft
lines on which the third line falls, making the ahernate angles, £c#
equal, must be in the same plane, without which they cannot be pa*
rallel, as is evident from the definitioa of parallel hnes. The only
editOT, I believe, who has remarked this omission, ts M . de Foix Duo
])E Cansalle, in his translation of the Elements published in 1566*
How it has escaped the notice of subsequent commentators is not ea-
sily explained, unless because they thought it of little importance toi
correct an error by which nobody was likely to be misled.

PROP. XXIX.

The strbject of parallel lines is one of the most difficult in the Ele*
ments of Geometry^ It hav accordingly been treated of in a great va*
riety of different ways, of which> perhaps, there is none that can be
said to have given entire satisfaction. The difficulty consists in eon*
verting the 27th and 28th of Euclid, or in demonstrating, that parallel
sbaight lines, or such as do not meet oiie another, when they meet a
third line, make the alternate angles with it equal, or, ^hich comes t(>
the same, are equally inclined to it, and make the exterior angle equal
to the interior and opposite. In order to demonstrate this proposition,.
Euclid assuined it as an Axiom, that " if a straight line kneet. two
*^ straight lines, so as to make the interior angles on th^ same side of il
** less than two right angles, these straight lines being continually pro*
^* diiced, will at length meet on the side on which the angles are that
^ are less than two right angles." This proposition, however, is n€>t
self-evident, and ought the less to be received without proof, that, as
Proclus has observed, the convene of it is a proposition that confesa^
edly requires to be demonstrated. For the converge of it is^ that two
straight lines which meet one another make the interior angles, with
any third line, less than two right angles ; or, in other words, that the
two interior angles of any triangle are less than two right angles,
which is thelTtb of the First Book of the Elements : and it should
seem, that a proposition can never rightly be taken for an Axiom, of
which the converse requires a demonstration.

The methods by which Geometers have attempted to remove this
blemish from the elements are of three kinds. 1. By a new definitioa
of parallel lines. 2. By introducing a new Axiom concerning paral-
lel lines, more obvious than Euclid's. 3. By reasoning merely front
the definition of parallels, and the properties of lines already demon-
strated without the assumption of any new Axiom.

1. One of the definitions that has been substituted for Euclid's is>
that straight lines are parallel, which preserve always the same dis*
lance from one anolfaer^ by the word distance M^g understood^ a per-

*•

NOTElS. • 301

pendicular drawn to one of the lines from any point whatever in thei
other. If these perpendiculars he every where of the same length, the
straight lines are called parallel. This is the definition given by
Wolfius, by Boscovich, and by Thomas Simpson, in the first edition of
his Elements. It is however a faulty (lefinition, for it conceab an
Axiom in it, and takes for granted a property of straight lines, that
ought either to be laid down as seli-evident, or demonstrated, if possi-
ble, as a Theorem. Thus, if from the three points A, B, and C of
the straight line AC, perpendiculars AD, BE, CF be drawn all equal
to one another, it is implied in the defini-
tion that the points D, E and F are in the
same straight line, which, though it be
true, it was not the business of ^he defini-
tion to inform us of. Two perpendiculars,
as AD and CF, are alone sufficient to de-
termine the position of the straight line DF, and therefore the defini-
tion ought. to be, *' that two straight lines are parallel, when there are
" two points in the one, from which- the perpendiculars drawn to the
" other are equal, and on the same side of it."

This IS the definition of parallels which M. D'Alembert seems to
prefer to all others ; but he acknowledges, and very justly, that it still
remains a matter of difficulty to demonstrate, that all the perpendicu-
lars drawn from the one of these lines to the other are .^qual. (jETncjf-
clopedie^ Art* Pardllele.) , ,

Another definition that has been given of parallels is, that they are
lines which make equal angles with a third line, toward the same parts, v
or such as make the exterior angle equal to the' interior and opposite.
Varignon, Bezout, and several other mathematicians, have adopted
this definition, which, it must be acknowledged, is a perfectly good

one, if it be understood by it, that the two lines called parallel, are
such as make equal angles with a certain third line, but not with any
line that falls upon- them* It remains, therefore, to be demonstrated,
I'hat if AB and CD make equal angles with GH, they will do so also
with any other line whatsoever. The definition, therefore, must be
thus understood^ That parallel lines are such as make equal angles^

302 • NOTE?.

with a certain third line, or, more simply, lined which are perpendica^
lar to a given Kne. It must then be proved, 1. That straight line^
which are equally inclined to a certain line or perpendicular to a cer-
tain line, must be equally inclined to all the other lines that fall upon
them ; and also, 2. That two straight lines which do not meet when
produced, must make equal angles with any third line that mee(s them;

The demonstration of the first of these propositions is not at all fa-
cilitated by the new definition, unless it be previously shown that all
the angles of a triangle are equal to two right angles.

The second proposition would hardly, be necessary if the new defi-
nition were employed ; for when it is required to draw a line that
shall not meet a given line, this is done by drawing a line that shalt
have the same inclination to a third line that the first or given line has.
It is known that lines so draiivn cannot meet. It would no doubt be
an advantage to have a definition that is not founded on a condition
purely negative.

2. As to the Mathematicians who have rejected Euclid's Axiom,
and introduced another in its place, it is not necessary that much
should be said. Clavius is one of the first in this class ; the Axiom
he assumes is, <* That a line of which the points are all equidistant
** from a certain straight line in the same plane with it, is itself a
** straight lino.'' This proposition he does not, however, assume alto-
gether, as he gives a kind of metaphysical proof of it, by which he
endeavours to connect it with Euclid's definition of a straight line,
with which proof at the same time he seems not very well satisfied.
His reasoning, after this proposition is granted (though it ought not to
1)0 granted as an Axiom), is logical and conclusive, byt is prolix and
operose, so as to leave a strong suspicion that the road pursued is by
no means the shortest possible.

The method pursued by Simson, in his Notes in the First Book of
Euclid, is not very difierent from that of Clavius. He assumes this
Axiom, " That a straight line cannot first come nearer to another
*' straight line, and then go farther from it without meeting it.'^

S Notes, dz;c. English Edition. ) By coming nearer is understood, con-
brmably to a previous definition, the diminution of the perpendiculars
drawn from the one line to the other. This axiom is more readily
assented to than that of Clavius, from which, however, it is niot very
dififerent : but it is not very happily expressed^ as the idea not merely
of motion, but of time, seems to be involved in^the notion of first com-
ing nearer, and then going farther off. Even if this inaccuracy is
passed over, the reasoning of Simson, like that of Clavius, is prolix,
and evidently a circuitous method of coming .at the truth.

Thomas Simpson, in the second edition of his Elements, has pre^
sented this Axiom in a simpler form. *< If two points in a straight
«< line are posited at unequal distances from another straight liner
<' in the same plane, those two lines being indefinitely produced ovt
^' the side of the least distance will meet one another."

By help of this Axiom it is easy to prove, that if two straight linea

NOTES.

dOS

JLB, CD are parallel, the perpendiculars to the one, terminated by the
other, are all equal, and are also perpendicular to both the parallels.
That they are equal is evident, otherwise the lines would mqet by
the Axiom. That they are perpendicular to both, is demonstrated
thus :

If AC and BD, which are perpendicular Vo AB, and equal to one
another, be not also perpendicular to
CD, from C let CE be drawn at right an-
gles to BD. Then, because AB and C£
are both perpendicular to BD, they are
parallel, and therefore the perpendiculars
AC and BE are equal. But AC is equal A.
to ISD, (by hypothesis,) therefore BE and BD are equal, which is im-
possible ; BD is therefore at right angles to CD.

Hence the proposition, that <' if a straight line fall on two parallel
*^ lines, it makes the alternate angles equal,*' is easily derived. Let

PH and QE be perpendicular to CD, then they will be parallel to on^
another, and also at right angles to AB, and therefore FG and HE
are equal to one another, by the last proposition. Wherefore in the
triangles EFG, EFH, the sides HE and EF are equal to the sidea
GF and FE, each to each, and also the third side HF to the third
side EG, therefore the. angle HEF is equal to the angle EFG, and
they are alternate angles. Q. E. D.

This method of treating the doctrine of parallel lines is extreme-
ly plain and concise, and is perhaps as good as any that can be fol-
lowed, when*a new Axiom is assumed. In the text above, I have,
however, followed a different method, employing as an Axiom, "That
'' two straight lines, which cut one another, cannot be both parallel
** to the same straight line." This Axiom has been assumed by
others, particularly by Ludlam, in his very useful little tract, entitled
Rudiments of Malhematics.^ ^

It is a proposition readily 'enough admitted as self-evident, and
leads to the demonstration of Euclid's 29th Proposition, even with
more brevity than Simson's.

3. All the methods above enumerated leave the mind somewhat dis-
satisfied, aj9 we naturally expect to discover the properties of parallel
lines, as we do those of other geometric quantities, by comparing the
definition of those linesi with the properties of straight lines already

804 NOTES.

known. The most ancient writer who appears to have attempted to
do this is Ptolemy the astronomer, who wrote a treatise expressly on .
the subject of Parallel Lines. Proclus has preserved some account
of this work in the Fourth Book of his commentaries : and it is cu-
rious to observe in it an argument founded on the principle which is
known to the modems by the name of the sufficient reason.

To prove, that if two parallel straight lines, AB and CD, be cut by
a third line EF, in Q and H, the two interior angles AGH, CHG will

be equal to two right angles, Ptolemy reasons thus : If the angles
AGIJ, CHG be not equal to two right angles, let them, if possible, be
greater than two right angles : then, because the lines AG and CH
are not more parallel than the lines 6G and DH, the angles BGH,
DHG are also greater than two right angles. Therefore, the four
angles AGH, CHG, BGH, DHG are greater than four right angles ;
and they are also equal to four right angles, which is absurd. In the
same manner it is shewn, that the angles AGH, CHG cannot be less
than two right angles. Therefore they are equal to two right angles.

But this reasoning is certainly inconclusive. For why are we to
suppose that the interior an^es which the parallels make with the
line cutting them, are either in every case greater than two right an-
gles, or in every case less than two right angles? For any thing that
we are yet supposed to know, they may be sometimes* greater than
two right angles, and sometimes less, and therefore we are not enti«
tied to conclude, because the angles AGH, CHG are greater than two
right angles, that therefore the angles BGH, DHG are also necessa-
rily greater than two right angles. It may safely be asserted, there-
fore, that Ptolemy has not succeeded in his attempt to demonstrate the
properties of parallel lines without the assistance of a new Axiom.

Another attempt to demonstrate the same proposition without the
assistance of a new Axiom has been made by a modern geometer,
Franceschini, Professor of Mathematics in the University of Bologna,
in an essay, which he entitles, La Teoria deUe paraikle HgarosamerUe
dimmstrataj printed in his Opuscdi Mathenuaid^ at Bassano in 1787.

The difficulty is there reduced to a proposition nearly the same
with this, That if BE make an acute angle with BD, and if D£ be

NOTESv

iperpeQdhcular to BD at any
point, BE and DE, if produc
«d, will meet. To demonstrate
this, it is supiposed, that BO,
BC are two parts taken in BE«
of which BC is greater than
BO, and that the perpendicu-
lars ON, CL are drawn to BD^
then shall BL be greater than
BN. For, if not, that is, if the
perpendicular CL falls either at
N, or between B and N, as at F j
in- the first of these cases (he angle CNB is equal to the angle ONB>
because they are both right angles, which is impossible ; and, in thib
second, the two angles CFN, CNF of the triangle CNF, exceed two
right angles. Therefore^ adds our author, since, as BC increases,
BL also increases, and since BC may be increased without limit, so
BL may become greater than any given line, and therefore may
be greater than BD ; wherefore, since the perpendiculars to BD from
points l^eyond D meet BC^ the perpendicular from D necessarily
meets it. Q. E. D.

Now it will be found, on examination, that tliis reasoning Is no
more conclusive than the preceding. For, unless it be proved, that
whatever multiple BC is of BO, the same i^ BL of BN, the indefinite
increase of BC does not necessarily imply the indefinite increase of
BL, or that BL may be made to exceed BD. On the contrary, BL
may always increase, and yet may do so in such a manner as never
to exceed BD : In order that the demonstration should be conclusive^
it would be necessary to shew, that when BC increases by a part
equal to BO, BL increases always by a part equal to BN ; but to do
this will be found to require the knowledge of those very properties
i)f parallel lines that we are seeking to demonsftrate.

Legendre, in his Elements of Geometry, a work entitled to ihe
highest praise, for elegance and accuracy, has delivered the doctrine
of parallel lines without any new Axiom. He has done this in two
dififerent ways, one in the text, and the other in the notes. In the
former he has endeavoured to prove, independently of the doctrine of
parallel lines,.that all the angles of a triangle are equal to two right
angles ; from which proposition, when it is once established, it is not
difficult to deduce every ^thing with respect to parallels. But, (hough
his demonstration of the property of triangles just mentiuned is quit©
logical and conclusive, yet it has the fault of being long and indirect,
proving first, that the three angles of a triangle cannot be greater than
two right angles, next, that they cannot be less, and doing both by
reasoning abundantly subtle^ and not of a kind readily apprehended
by those who are only beginning to study the Mathematics.

The demonstration which he has given in the notes is extremely in*

iM NOTES. '

genious, and proceeds on this very simple and undeniable Axiom, that
we cannot compare an angle and a line, as to magnitude, or cannot
have an equation of any sort between them. This truth is involved in
the distinction between homogeneous and hetervgeneous quantities,
(Eac. v. def. 4.), which has long been received in Geometry, but led
only to negative conseauences, fill it fell into the hands of Legendre.
The proposition which he deduces from it is, that if two angles of one
triangle be equal to two angles of another, the third angles of these
triangles are also equal. For, it is evident, that when two angles of a
triangle are given, and also the side between them, the third angle is
thereby determined ; so that if A and B be any two angles of a trian-
gle, P the side interjacent, and C the third angle, C is determined, as
to its magnitude, by A, B and P ; and, besides these, there is no other
quantity whatever which can aflf^ct the magnitude of C This is plain,
because if A, B and P are given, the triangle can be constructed, all the
triangles in which A, Band Pare the same, being equal to one another.

Bnt of the quantities by which C is determined, P cannot be one ;
for if it were, then C must be erfunctionof the quantities A,'B, P ;
that is to say, the value of C can be expressed by some combination of
the quantities A, B and P. An equation, therefore, may exist be-
tween the quantities A, B, C, and P ; and consequently the value of
P is equal to some combination, that is, to some function of the quan-
tities A, B and C ; but this is impossible, P being a line, and A, B, C
being angles ; so that no function of the first of these quantities can
be equal to any function of the other three. The angle C must there-
fore be determined by the angles A and B alone, without any regard
to the magnitude of P, the side interjacent. Hence in all triangles
that have two angles in one equal to two in another, each to each,
the third angles are also equal.

Now, this being demonstrated, it is easy to prove that the three
angles of any triangle are equal to two right angles.

Let ABC be a triangle right angled at A, draw AD perpendicularto
EC. The triangles ABD, ABC have the J^

angles BAC, BDA right angles, and the
s angle B common to both ; therefore by
what has just been proved, their third an-
gles BAD, BCA are also equal. In the
same way it is shewn, that CAD is equal

to CBA ; therefore the two angles, BAD -n r% r^

CAD are equal to the two BCA, CBA ; but *" ^ ^

BAD+CADis equal to a right angle, therefore the angles BCA, CBA
are together equal to a right angle, and consequently the three angles
of the right angled triangle ABC are equal to two right angles.
^ And since it is proved that th^ oblique angles of every right angled
triangle are equal to two right angles, and since every-triangle may
be divided into two right angled triangles, the four oblique angles of
which are equal to the three angles of the triangle, therefore the three
angles of every triangle are equal to two right angles. Q. E. Du

* NOTES. 80»

Though this method of treating the subject is strictly demonstrative^
yety as the reasoning in the first of the two preceding demon strationq
is not perhaps sqfficiently'simple to be apprehended by those just en-
tering on matheniatical studies^ X' shall submit to the reader another
method, not liable to the same objection, which I know, from expe*
rience, to be of use in explaining the Elements. It proceeds, like that
of the French Geometer, by demonstrating, in the ^rst'place, that thQ
angles of any triangle are together «qual to two right angles, and de->
ducing from thence, that two lines, which make with a third line thQ
interior angles, less than two right angles, must meet if produced. The
reasoning used to demonstrate the first of these propositions may be
objected to by some as involving the idea of motion, and the transfer^
ence of a line from one place to another. This, however, is no more
than Euclid has done himself on some occasions; and when it furnish*
es so short a road to the truth as in the present instance, and does nof
impair the evidence of the conclusion, it seems to be in no respect
inconsistent with the utmost rigour of demonstration. It is of im-
portance in explaining the Elements of Science, to connect truths by
the shortest chain possible ; and till that is done, we can npver con*
sider them as being placed in their natural order. The reasoning in
the first of the following propositions is so simple, that it seems hard-
ly susceptible of abbreviation, and it has the advantage of connect<*
ing immediately two truths so much alike, that one might conclude^
even from the bare enunciations, that they are but different cases of
the. same general theorem, viz. That all the angles about a i>oipt^
and all the exterior angles of any rectilineal figure, are constantly of
the same magnitude, and equal to four right angles.

DEFINITION.

If, while one extremity of a straight line re-
mains fixed at A, the line itselPturns about that
point from the position AB to the position AC, it
is said to describe the angle BAG contamed by
the line AB and AC.

Cor. If a line turn about a point from the position AC till it come
into the position AB again, it describes angles which' are together
equal to four right angles. This is eyident from the second Cor. to
the 15th.

PROP. I.

All the exterior angles of any rectilineal figure are together ejgf<(al
to four right angles.

1. Let the rectilineal figure be the triangle ABC, ofvUich the ex*
terior angfes are DC A, FAB, GBC ; these angles ard together equal
to four right angles.

SOB

NOTES.

Let the fine CD, placed in the direction of BC prodaccd, torn
about the point C till it coincide with CE, a part of the aide CA, and
have described the exterior angle DCE or DC A. Let it then be car-
ried along the line CA, till it
be in the position AF, that iff,
in thedirection of C A produc-
ed, and the point A remaiir*
ing fixed,' let it turn about A
till it describe the angle FAB,
and coincide with a part of the
line AB. Let it next be car-
ried along A^ till it come into
the position BG, and by turn-
ing about B, let it describe
the angle 6BC, so aa to coin-
cide with a part of BC. ^ Last-
ly, Let it be carried along BC
till it coincide with CDr ita
first position. Then, because
the line CD has turned about
one of its extremities till it
has come into the position CD again, it has by the corollary to the
abore definition described angles which are together equal to four
right an^es ; but the angles which it has described are the three
exterior angles of the triangle ABC, therefore the exterior angles of
ihe triangle ABC are equaFto four right angles.

2, If the rectilineal figure have any number of sides, the proposi-
tion is demonstrated just as in the case of a triangle. ^ Therefore all
the exterior angles of any rectilineal figure are together equal to four
right angles. Q. £. D.

CoRi. 1. Hence, alL the fnterror angles of any triangle are equal to
two right angles.. For all the angles of the triangle, both, exterior
and interior, are equal to six right angles, and the exterior being,
equal to four right angles, the interior are equal to* two right angles..

CoB. 2. An exterior angle of any triangle is equal to the two inte-
rior and opposite, or the angle DCA is^ equal to the angles. CAB,
ABC. For the angles CAB, ABC, BCA are equal te two right an--
glea; and the angles ACD, ACB are also (1^. 1.) eqiual to two right
angles ; therefore the three angles CAB, ABC, BCA are equal to
the two ACD, ACB ; and taking ACB from, both, the angle ACD ia
equal to the two angles CAB^ ABC.

Cos. 3.. The interior angles of any rectilineal figure are equal to
fwit^ as many right angles as the figure has sides, wanting four. For
all the {kfdgles exterior and interior are equal to twice as many right
angles as th^ figure has sides; but the exterior are equal to four right
angles ; therefove the interior are equal to twice aa many right aib-
gles as the figure bma sides, wanting four.

NOTES. 309

PROP. II.

Two straight lines, which make with a third line the interior angles
on the same side of it less than two right angles, will meet on that
side, if produced far enough.

Let the straight lines AB, CD, make with AC the two angles BAC,
DCA less than two right angles ; AB and CD will meet if produced
toward B and D.

In AB take AF=AC ; join CF, produce B A to H, and through C
draw CE, making the angle ACE equal to the angle CAH.

Because AC is equal to AF,the angles AFC, ACF are also equal (5^
1.) ; but the exterior angle HAC is equal to the two interior and op-
posite angles ACF, AFC, and therefore it is double of either of them,
as of ACF. Now ACE is equal ^o HAC by construction, therefore-
ACE is double of ACF, and is bisected by the line CF. In the same
manner, if FG be taken equal to FC, and if CG be drawn, it may be
shewn that CG bisects the angle ACE, and so on continually. But if
from a magnitude, as the angle ACE, there be taken its haJf, and frbm
the remainder FCE its half FCG, and from the remainoer GCE its
half, &c. a remainder will at length be found less than the given an-
gle DCE.*

Let GCE be the angle, whose half ECR is less than DCE, then a
straight lin^ CK is found, which falls between CD and C£, but never-
theless meets the line AB in K. Therefore CD, if produced, must
meet AB in a point between G and K. Therefore, &c. Q. E. D.
This demonstration is indirect ; but this proposition, if the defini-
tion of parallels Vere changed, as suggested at p. 302, would not be
necessary ; ajid the proof, that lines equally inclined to any one line
must be so to every line, would follow directly from the angles of a
triangle being equal to two right angles. The doctrine of parallel
lines would in this manner be freed from all difficulty.

PROP. m. 29. 1. Euclid.

If a straight line fall on two parallel straight lines, it makes the al-
ternate angles equal to One another ; the exterior equal to the interior

- * Prop. 1. 1. Sap. The reference of (his proposition involves nothing inconsistent with
fi^ood reasoning', as the demonstration of it does not de|tend on any thing that has gone ba«
ibroi so tiiat U m^ be iotrodacedjn any part of the ElemeiiiB.

810 NOTES.

and opposite on the same side ; and likewise the two interior angles,
on the same side equal to two right angles.

Let the straight line EF fall
on the parallel straight lines £'

AB, CD ; the alternate' an-
gles A G Hj GHD are equal, the
exterior angle EGB is equal
to the interior and opposite
GHD ; and the two interior
angles BGH, GHD are equal
to two right angles.

For if AGH be not equal to
GHD, let it be greater, then
adding BGH to both, the an-
gles AGH, HGB are greater than the angles DHG, RGB. But AGH,
HGB are equal to two right angles (13. 1.); therefore BGH, GHD are
less than two right angles, and therefore the lines AB, CD will meet,
by the last proposition, if produced toward B and D. But they do nt>t
meet, for i\thy are parallef by hypothesis, and therefore the angles
AGH, GHD are not unequal, that is, they are equal to one another.

Now the angle AGH is equal to EGB, because these are vertical,
and it has also been shewn to be equal to GHD, therefore EGB and
GHD are equal. Lastly, to each of the equal angles EGB, GHD add
the angle BGH, then the two EGB, BGH are equal to the two DHG,
BGH. But EGB, BGH are equal to two right angles, (13. 1.), there-
fore BGH, GHD are also equal to two right angles. Therefore, ^c.
Q. E. D.

The following proposition is placed here, because it is more con-
nected with the First Book than with any other. It is useful for ex-
plaining the nature of Hadley.'s sextant ; and, though ioyolved in the
explanations usually given of that instrument, it has not, I believe,
been hitherto considered as a distinct Geometrical Proposition, though
very well entitled to be so on account of its simplicity and elegance,
as well as its utility.

THEOREM. ^

If an exterior angle of a triangle be. bisected, and also one of the
interior and opposite, the angle contained by the bisecting lines is
equal to half the other interior and opposite angle of the triangle.

Let the exterior angle ACD of the triangle ABC be bisected by
the straight lineCE, and the interior and opposite ABC by the straight
line BE, the angle EEC is equal to half the angle 6AC.

The line CE, BE will meet ; for since the angle ACD is greater than
ABC, the half of ACD is greater than the half of ABC, that is^ ECD

NOTES.

811

id greater than EBC ;
add ECB to both, and
the two angles ECD,
ECB are greater than
EBC, ECB. But ECD,
ECB ai*e equal to two
right angles ; therefore
ECB, EBC are less
than two right angles,
and therefore the lin^s
C£, BE must meet on

the same side of BC on which the triangle ABC ib. Let them meet
in E.

Because DCE is the exterior angle of the triangle BCE, it is equal
to the two angiesCBE, BEC, and therefore twice the angle Dp£,that
is, the angle DCA is equal to twice the angles CBE and BEC; But
twice the angle CBE is equal to* the angle ABC, therefore the angle
D Ac is equal to the angle ABC, together with twice the angle BEC ;
and the satneangle DCA being the exterior angle of tlie triangle ABC,
is equal to the two angles ABC, CAB, wherefore the two angles ABC,
CAB are equal to ABC and twice BEC. Therefore, taking away
ABC from both, there remains the anglcCAB equal to twice the angle
BEC, or BEC equal to the half of BAC, Therefore, <&c. ft. E. D.

BOOK II.

The Demonstrations of this Book are no otherwise changed than
by introducing into them some characters similar to those of Algebra^
which is always of great use where the reasoning turns on the addi-
tion or subtraction of rectangles. To Euclid's demonstrations, others
are sometimes added, serving to deduce the propositions from the
fourth, without the assistance of a diagram.

PROP. A and B.

These Theorems are added on account of their great use in geo-
metry, and their close connection with the other propositions which are
the subject of this Book. Prop. A is an extension of the 9th and 19th.

BOOK. III.

DEFINITIONS.

The definition which Euclid makes the first of this Book is that of
equal cirdes; which he defines to be ** those of which the diameters

613 NOTES*

*< are equal." This is rejected from among the definitions, as being
a Theorem, the truth of which is proved by supposing the circles ap*
plied to one another, so that their centres may coincide, for the
whole of the one must then coincide with the whole of the other.
The converse, viz. That circles which are equal have equal diame-
ters, is proved in the same way.

The definition of the angle of a segment is also omitted, because
it does not relate to a rectilineal angle, but to one understood to be
contained between a straight line and a portion of the circumference
of a circle. In like manner, no notice is taken in the 16th proposi-
tion of the angle comprehended between the semicircle and the dia«
meter, which is said by £uclid to be greater than an acute rectili-
neal angle. The reason for these omissions has already been assign-
ed in the notes on the fifth definition of the first Book.

PROP. XX.

It has been remarked of this demonstration, that it takes for grant-
ed, that if two magnitudes be double of two others, each of each, the
sum or difference of the first two is double of the sum or difiTerence
of the other two, which are two cases of the 1st and 5th of the 5th
Book. The justness of this remark cannot be denied ; and though the
cases of the Propositions^ here referred to are the simplest of any, yet .
the truth'bf them ought not in strictness to be assumed without proof.
The proof is easily given. Let A and^B, C and D be four magnitudes,
such that A=2C, and B=aD ; then A+B=2(C + D). For since
A=C+C, and B— D+D, adding equals to equals, A+B=(C+D)
.KC4-D)=2(C+D). So also, if A be greater than B, and therefore
O greater than D, since A=^C 4-0, and B^^D+D, taking equals from
equals, A— B=(C— D)+(C— D) that is. A— B=2(C— D).

BOOK V.

The subject of proportion has been treated so differently by those
who have written on elementary geometry, and the method which
Euclid has followed has been so often, and so inconsiderately cen-
sured, that in these notes it will not perhaps be more necessary to ac*
count for the changes that I have made, than for those that I have
not made. The changes are but few, and relate to tl^e language, not
to the essence of the demonstrations ; they will be explained after
some of the definitions have been particularly considered.

DEF. III.

The definition of ratio given here has been greatly extolled by some
authors ; but whatever value it^may have in the eyes of a metaphysi-
cian, it has but little in those of the geometer^ because nothing coneern*

Notes. • ais

kig die properties of ratios, can^be deduced from it. Dt, Barrow has
very judiciously remarked concerning it, *^ that Euclid had probably
*^ no'other design in making this definition, than to give a general sumw
** mary idea of ratio to beginners, by premising this metaphysical defi»
^^ nition to the more accurate definitions of .ratios that aire equal to on0
*^ another, or one of which is greater or less than the other ; I call
^* it a metaphysical, for it is not properly a mathematical definition,
** since nothing in mathematics depends on it, or is deduced, nor, as I
** judge, can be deduced, from it." (Barrow's Lectures, Lect. S.)
Dr. Simson thinks the definition has been added by some unskilful edi.
tor ; but there is no ground for that^ supposition, other than what arf-
ses from the deiinition being of no use. We may, however, well
enough imagine, that a certain idea of order and method induced Eu-
clid to give some general definition of ratio before he used the term
in the definition of equal ratios.

DEF. IV.

This deiinition is a little altered in the expression ; Euclid has it)
that *^ magnitudes are said to have a ratio to one another^ when the
^' less can be multiplied so as to exceed the greater."

DEF. V-

One of the chief obstacles to the ready understanding of the 5th
Book of Euclid, is the difficulty that most people find of reconciling the
idea of proportion which they have already acquired, with the account
of it that is given in this definition. Our first ideas of proportion, or
of proportionality, are got by trying to compare together the magni-
tude of external bodies ; and though they be at first abundantly vague ~
and incorrect, they are usually rendered tolerably precise by the
study .of arithmetic ; from which we learn to call four numbers pro-
portionals, when they are such that the quotient which arises from di-
viding the first by the second, (according to the common rule for divi-
sion), is the same with the quotient that arises from dividing the third
by the fourth. '

Now, as the operation of arithmetical division is applicable as rea-
dily to any two magnitudes of the same kind, as to two numbers, the
notion of proportion thus obtained may be considered as perfectly
general. For, in arithmetic, after finding how oflen the divisor is
contained in the dividend, we multiply the remainder by 10, or 100,
or 1000, or any power, as it is called, of 10, and proceed to inquire
how oft the divisor is contained in this new dividend ; and„ if therQ
be any remainder, we go on to multiply it by 10, 100, &;c. as before,
and to divide the product by the original divisor, and so on, the divi-
sion sometimes terminating when no remainder i^ left, and sometimes
going on ad infinitum, in consequence of a remainder being left at each
operation. Now, this process may easily be imitated with any two

'40

814 NOTES.

magnitudes A and B, providing they be of the same kind, of* such thcU
the one can be multiplied so as to exceed the other. For, suppose
that B is the least of the two ; take B out of A as oft as it can be found,
and let the quotient be noted, and also the remainder, if there be any ;
multiply this remainder by 10, or 100, &e. so as to exceed B, and let
B be taken out of the quantity produced by this multiplication as oft
as it can be found ; let the quotient be noted, and also the remain-
der, if there* be any. Proceed with this remainder as before, and so
on continually ; and it is evident, that w6 have an operation that is
applicable to all magnitudes whatsoever, and that may be performed
with respect to any two lines, any two plane figures, or any two solids,
&c.

Now, when we have two magnitudes and two others, and find that
the first divided by the second, according to thi» method, gives the ve-
ry same series of quotients that the third does i^^hen divided by the
fourth, we say of these magnitudes, as we did of the numbers above
described, that the first is to the second as the third to the fourth.
There are only two more circumstances necessary to be considered,
in order to bring us precisely to Euclid'^ definition.

First, It is known from' arithmetic, that the multiplication of the ^
successive remainders each of them by 10, is equivalent to multiply-
ing the quantity to be divided by the product of all those tens ; so
that multiplying, for instance, .the first remainder by 10, the second by
10, and the third by 10, is the same thing, with respect to the quo-
tient, as if the quantity to be divided had been at first multiplied by
1000 ; and therefore, our standard of the proportionality of numbers
may be expressed thus : If the first multiplied any number of times
by 10, and then divided by the second, gives the same quotient as
when the third is multiplied as often by 10, and then divided by the
fourth, the four magnitudes are proportionals.

Again, it is evident, that there .is no necessity in these multiplica-
tions for confining ourselves to 10, or the powers of 10, and that we
do so, in arithmetic, only for the conveniency of the decimal notation ;
we may therefore use any multipliers whatsoever, providing we use
the same in both cases. Hence, we have this definition of propor-
tionals. When there ai:e four magnitudes, and any multiple whatsoever
of the first, when divided by the second, gives the same quotient with
the like multiple of the third, when diyided by the fourth, the fouc
magnitudes are proportionals, or the first has the same ratio to the se-^
cond that the third has to the fourth.

We are now arrived very nearly at Euclid's definition ; for, let A>
B, C, D be four proportionals, according to the definition just given,
and m any number ; and let the multiple of A by m, that is mA,-be
divided by B ; and first, let the quotient be the number n exactly, then
also, when mC is divided by D, the quotient will be n exactly. But
when mA divided by B gives n for the quotient, mA=nB by the nature
of divisioii, so that when mA=nB, wC=caD, which is one of the con-
ditions of Euclid's definition. ' .

NOTES. S15

Agajp, when m^. is divided by B, let the division not be exactly
performed, but let n be a whole number less than the exact quotient,
then nBZ.mA, or mA ynB ; and, for the same reason, mCyhD, which
is another of the conditions of Euclid's definition.

Lastly, when »iA is divided by B, let n be a whole number greater
than the exact quotient, then mAZnB, and because n is also greater
than the quotient of mC divided by Dj (which is the same with the
other quotient), therefore mCZnD.

Therefore, uniting ail these three conditions, we call A, B, C, D,
proportionals, when they are such, that if mAynB, mCynD ; if mA.
=nB, «iC=nD ; and if mAZwB, mC^hDy m and n being any num-
bers whatsoever. Now, this is exactly the criterion of proportionality
established by Euclid in the 5th definition, and is derived here by
generalizing the common and most familiar idea of proportion.

It appears from this, that the condition of mA containing B, whether
with or without a remainder, as often as mC contains -D, with or with-
out a remainder, and of this being the case whatever value be assigned
to the number m, includes in it all the three conditions that are men-
tioned in Euclid's definition ; and hence, that definition may be ex-
pressed a little more simply by saying, thtdfour magnitudes are pro^
jwrtionah, when any mtdtiple of the first contains the second, {with or
without remainder,) as oft as the same multiple of the third contains the
fourth. ' But, though this definition is certainly, in the expression,
more simple than Euclid's, it is not, as will be found on trial, so easily
applied to the purpose of demonstration. The three conditions which
Euclid brings together in his definition, though they somewhat embar-
rass the expression of it, have the advantage of rendering the demon-
strations more simple than they would otherwise be, by avoiding all
discussion about the magnitude of the remainder lefl, afler B is tak^n
out of m A as oft as it can be found. All the attempts, indeed, that have
been made to demonstrate the properties of proportionals rigorously,
by means of other definitions than Euclid's, only serve to evince the
excellence oT the method followed by the Greek Geometer, and his
singular address in the application of it.

The great objection to the other methods is, that if they are meant
to be rigorous, they require two demonstrations to every proposition,
one when the division of m A into parts equal to B can be exactly per-
formed, the other when it cannot be exactly performed n^hatever value
be assigned to m, or when A and B are what, is calM incommensura-
ble ; and this last case will generally be found ia require an indirect
demonstration, or a reductio Ctd absurdum,

M. D'Alembert, speaking of the doctrijx^ of proportion, in a dis-
course that contains many excellent observations, but in which he has
overlooked Euclid's manner of trep^ng this subject entirely; has the
following remark : " On ne pe^t demontrer que de cette manidre,
" (la reduction k absurde,) U plnpart des propositions qui regardent
" les incommensurables. L'id6e de I'ihfini entre au moins implicite-
" mens dans la notion de ces sortes de quantit^s ; et comme nous n'a-

S16 NOTES.

** ton0 qu'oQe icl6e negative de I'infini, on ne pent d^montrer di rectc-
** ment, et a priori, tout ce qui concerne I'infiai math^maiiqHe."
^Encyclopedic, mot G^omHrie.)

This remark sets in a strong and just light the difficulty of demon-
strating the propositions that regard the proportion of incommensura-
ble magnitudes, without having recourse to the reductio ad dbsurdum z
but it is suprising, that M . D' Alembert, a geometer no less learned
than profound, should have neglected to make mention oC Euclid's me**
thiKl, the only one in which the diiBculty he states is completely over-
come. It is overcome by the introduction of the idea of indefinitude, (if
I may be permitted to use the word), instead of the idea of infinity ; for
m and n, the multipliers employed, are supposed to be indefinite, or to ad-
mit of all possible values, and it is by the skilful use of Xhis condition that
the necessity of indirect demonstrations is avoided. In the whole of ge-
ometry, I know not that any happier invention is to be found ; and it is
worth remarking, that £uclid appears in another 6i his works to'tiEve
availed himself of the idea of indefinitude with the same success, viz.
in his books of Porisms, which have been restored by Dr. Simson,
and in which the whole analysis turned on that idea, as I have shown
at length in the Third Volume of the Transactions of the Royal So-
ciety of Edinburgh. The investigations of these propositions Were
founded entirely on the principle of certain noagnitudes admitting of
innumerable values ; and the methods of reasoning concerning them
seem to have been extremely similar to those employed in the fifth of
the Elements. It is curious to remark this analogy between the dif-
ferent works of the same author ; and to consider, that the skill, ia
the conduct of this very refined and ingenipus artifice, acquirpd ia
treating the properties of proportionals, may have enabled Euclid ta
succeed so well in treating the still more difficult subject of Porisms.

Viewing in this light Euclid's manner of treating proportion, I had
no desire to change any thing in the principle of hi& demonstrations*
I have only sought to improve the language of them, by introducing
a concise mode of expression, of the same nature with, that which we
nse in arithmetic, and in algebra. Ordinary language conveys the
ideas of the different operations supposed to be performed in these de-
monstrations so slowly, and breaks them down into so many parts, that
they make not a sufficient impression on the understanding* This in^
deed will generally happen when the things treated of are not repre-
sented to the senses by Diagrams, as they cannot be when we reasoa
concerning magniliMe in general, as in this, part of the Elements.
Here we ought certainVy to adopt the language of arithmetic or alge-
bra, which, by its shortnesj.> and the rapidity with which it places ob-
jects before us, mdkes up in tu best manner possible for being merely
a conventumal language,, and using symbols that have no resemblance
to the thmgs expressed by them. Su^h a language, therefore, I have
cndeavmired to introduce here ; and I ,m convinced, that if it shall
be found an improvement, it is the only one of which the fifth of Eu-
clid wiJl admit. In other respects I have followed Dr. Simson's «dir

NOTES« S17

tioD, to the accuracy of which it would be difficult to make any addi-
tion.

In one thing I must observe, that the doctrine of proportion, as laid
down here, is meant to be more general than in Euclid's Elements. It
is intended to include the properties of proportional numbers as well
as of all magnitudes. Euclid has not this design, for. he has given a
definition of proportional numbers in the seventh Book, very different
from that of proportional magnitudes in the fifth ; and it is not easy to
justify the logic of this manner of proceeding ; for we can never speak
of two numbers and two magnitudes both having the same ratios, un.
less the word ratio have in both cases the same signification. All the
propositions about proportionals here given are therefore understood
to be applicable to numbers ; and accordingly, in the eighth fiook^
the proposition that proves equiangular parallelograms to be in a ra-
tio compounded of the ratios of the numbers proportional to their sides,
is demonstrated by help of the propositions of the fiflh Book.

On account of this, the word quantity ^ rather than Toagnitude^ ouglit
in strictness to have been used in the enunciation of these propositions^
because we employ the word Quantity to denote not only things ^x*
tended, to which alone we give the name of Magnitude, but also num-
bers. It will be sufiicient, however, to remark, that all the proposir ^
tions respecting the ratios of magnitudes relate equally to all things of
which multiples can be taken, that is, to all that is usually expressed
by th^ word Quantity in its most extended signification, taking care al-
ways to observe, thfit ratio takes place only among like quantities^
(See Def. 4.)

DEF. X.

The definition of compound ratio was first given accurately by Dr*
Simson ; for, though Euclid used the term, he did so without definfng
it. 1 have placed this definition before those of dwpLieate and trtph'
cote ratioj as it is in fact more general, and as the relation of all the
three definitions is best seen when they are. ranged in this order. It
is then plain, that two equal ratios compound a ratio duplicate of either
of them ; three equal ratios, a ratio triplicate of either of them, &c.

It was justly observed by Dr. Simson, that the expression, compound
ratio, is introduced merely to prevent circumlocution, and for the 'sake '
principally of enunciating those propositions with conciseness that
are demonstrated by reasoning ex aquo, that is, by Reasoning froin the
22d or 23d of this Book. This will be evident to any one who^jcon-
siders cTarefully the Prop. F. of this, of the 28d of the 6th Book.

. An objection which naturally occurs to the use of the term compound
ratio, arises from its not being evident how the ratios described in
the definition determine in any way the ratio which they are said'
to compound,. since the magnitudes compounding them are assumed
at pleasui'e. It may be of use for removinfg this difficulty, to state .
the matter as follows : if there be any number of ratios (among mag-
nitudes of the same kind) such that the consequent of any of them is

8ie NOTES.

the antecedent of that which immediately follows, the first of the ati-
tecedeots has to the last of the consequents a ratio which evidently
depends on the intermediate ratios, because if they are determined,
it is determined also ; and this dependence of one ratio on all the other
ratios, is expressed by saying that it is compounded of them. Thus,
A R C D

^^D> ri* f\i en ^o ^^y series of ratios, such as described above, the
D U 11 Jci

A ' A H

ratio =y or of A to E, is said to be Compounded of the ratio ^ r^, &c.

A ' A B

The ratio ^r, ■« evidently determined by the ratios ^ ^^ &c, because

if each of the latter is fixed and invsjriable, the former cannot change.
The exact nature of this dependence, and how the one thing is deter-
mined by the other, it is not the busines~d of the definition to explain,
but merely to give a name to a relation which it may be of importance
to consider more attentively.

BOOK Vi.

DEFINITION 11.

This definition is changed from that o£ redprocdl Jigures, which was
of no -use, to one that corresponds to the language .used in the 14th and
15th propositions, and in other parts of geometry.

PROP. XXVII, XXVIII, XXIX.

As considerable liberty has been taken with these propositions, it
is necessary that the reasons for doing so should be explained. In the
fint place, when the enunciations are translated literally from the
Greek, they sound very harshly, and are, in fact, extremely obscure.
The phrase of applying to a straight line, a parallelogram deficient, or
exceeding by another parallelogram, is so elliptical, and so little analo-
gous to ordinary language, that there could be no doubt of the proprie-
ty of at least changing the enunciations.

It next occurred, that the Problems themselves in the 28th and 29th
propositions are proposed in a more general form than is necessary in
an elementary work, and that, therefore, to take those cases of them
that are the most useful, as they happen to be the most simple, must
be the best way of accommodating them to the capacity of the learner.
The problem which Euclid proposes in the 28th is, '< To a given
<< straight line to apply a parallelogram equal to a given rectilineal
** figure, and deficient by a parallelogram similar to a given parallelo.
" gram ;" which may be more intelligibly enunciated thus : ** To cut
«> a given line, so that the parallelogram which has in it a given angle,
*^ and is contained under one of the sejgments of the given line, and a

NOTES. 819

^' sUaight line which has a given ratio to the other segment^ may be
*^ equal to a given space ;" instead of which problem I have substitut-
ed this other ; *^ To divide a given straight line so that the rectangle
<< under its segments may be equal to a given space." In the actual
solution of problems, the greater generality of the former proposition
is an advantage more apparent than real, and is fully compensated by
the simplicity of the latter, to which it is always easily reducible.

The same may be said of the 2dth, which Euclid enunciates thus :
'* To a given straight line to apply a parallelogram equal to a given
*^ rectilineal figure exceeding by a parallelogram similar to a given pa-
** rallelogram." This might be proposed otherwise : <* To produce a
*^ given line, so that the parallelogram having in it a given angle, and
** contained by the whole line produced, and a straight line that has a
** given ratio to the part produced, may be. equal to a given rectilineal
*^ figure." Instead of this, is given the following problem, more sim-
ple, and, as was observed in the former instance, very little less gene-
ral. '^ To produce a given straight line, so that the rectangle contain-
*^ ed by the segments, between the extremities of the given line, and
*^ the point to which it is produced, may be equal to a given space."

PROP. A, B, C, &c.

Nine propositions are added to this Book on account of their utility
and their connection with this part of the Elements. T^e first four
of them are in Dr. Simson's edition, and among these Prop. A is given
immediately after the third, being, in fact, a second case of that pro-
position, and capable of being included with it, in one enunciation.
Prop. D. is remarkable for being a theorem df Ptolemy the Astpono*
mer, in his Msyaikn 2uvra|i(, and the foundation of the construction of
his trigonometrical tables. Prop. E is- the simplest case of the fbr«
mer ; i( is also useful in trigonometry, and, under another fomiy was
the 07tti, or, in some editions, the 94th of Euclid's Data. The pro^
positions F and G are very useful properties of the circle, and are
taken from the Lad Plani of Apollonius. Prop. H is a very remaik-
able property of the triangle ;' an4 K is a proposition which, though it
has been hitherto considered as belonging particularly to trigonome*
try, is so often of use in other parts of the mathematics, that it may
be properly ranked among the elementary theorems of Geometry;

820 NOTES.

SUPPLEMENT

BOOK I.

PROP. V. and VI, &c.

Ths demonstratioas of the 5th and 6th propositions require the
laeihod of exhaustiona, that is to say, they prove a certain pra**
perty to belong to the circle, because it belongs to the rectilineal
figures inscribed in it, or described a'bout it according to a certain lavtr,
in the case when those figures approach to the circles s6 nearly as not
to fall short of it or to exceed it, by any assignable difference, '{"his
principle is general, and is the only one by which we can possibly
compare curvilineal wjth rectineal spaces, or the length of curve
liaea witli the length of straight lines, whether we follow the methods
of the ancient or of the modern geometers. It is therefore a great
injustice to the latter methods to represent them as standing on a foun-
dation less secure than the former : they stand in reality on the same,
and the only difference is, that the application of the principle, com-
moa to them both, is more general and expeditious in the one case
than in the other. This identity of principle, and affinity of the
methods used in the elementary and the higher mathematics, i( seems
the more necessary to observe, that some Jearned tnathematicians have
appeared not to be sufficiently aware of it, and have even endeavoured
to demonstrate the contrary. An instance of this is to be met with
in the preface of the valuable edition of the works of Archimedes,
lately printed at Oxford. In that preface, Torelli, the learned com-
mentator, whose labours have done so much to elucidate the writings
of the Greek Geometer, but who is so unwilling to acknowledge the
merit of the modern analysis, undertakes to prov9,that it iis impossible,
from the relation which the rectilineal figures inscribed in, and cir-
cumscribed about, a given curve have to one another, to conclude
any thing concerning the properties of the curvilineal space itself, '
except in certain circumstances which he hast not precisely described.
With this view he attempts to show, that if we are to reason from the
relation which certain rectilineal figures belonging to the circle have
to one another, notwithstanding that those figures may approach so
tfiear to the circular spaces within which they are inscribed, as not to
differ from them by any assignable magnitude, we shall be led into
error/ and shall seem to prove, that the circle is to the square of its

NOTES. SUPPL, BOOK L 831

fjiametor exactly aa 3 to 4. Now, as this is a conclusion which the
discoveries of Archimedes himself prove so clearly to be false, To-
^ reUi argues, that the principle from which it is deduced must be false
also ; and in this he would no doubt be right, if his former cooclu-
aion had been fairly drawn. But the truth is, that a very gross para-
logism is to be found in that part of his reasoning, where he m^kea
a transition from the ratios of the small rectangles, inscribed in the
circular spaces, to the ratios of the sums of those rectangles, or of
the whole rectilineal figures. In doing this, he takes for grai^ted a
proposition, which, it is wonderful, that one who bad studied geome-
try in the school of Archimedes, should for a moment have supposed
to be true. The propositioa is this : If A, B, C, D, E, F, be any
number of magnitudes, and a, (, e, dy e^f, as many others ; and if
A : B : : a : 6,
C : D : : c : «2,

E : F : : e :/, then the sum of A, C and E will be to the sum of B,
P and F, as the sum of a, e and e, to the sum of $, d and/, or A +
C+E:B + D + F:: a+c+e : b+d-hf- Now, this proposition,
whieh Torelii supposes to be perfectly general, is not true; except in
two cases, Tiz« either first, when A : C : : a : c, and

A : E : : a : 6 ; and consequently,
B iD :i b; d, and
B : F : : 6 :/ ; or, secondly^ when
441 the ratios of A to B, C to D, E to F, fi^c. are equal to one another.
To -demonstrate this, let us suppose that there are (oiyc magnitudes,
and four others,

thus A : B : : a : .5, and

C : D : : c : i2, then we cannot have
A -f ^ • ^ 4* ^ 2 > ^"^^ • h^df unless either A : C : : a : c, and fi :
D : :b : d ; or A : C : : b : d, and consequently a : b : :c : d.

Take a magnitude K, such that a.:c : : A : K, and another L, such
ithat b: d : :B ',1j; and suppose it true, that A + C :
B+D : : a-^-c : b-i-d* Then, because by inversion ;
.K : A : : c ; a, and, by hypothesis, A : B : : a : 6, and
also B : L : : 6 : <2, ex equo, K : L : : c : d ; and con-
sequently, K : L : : C : D.

Again, because A : K : : a : c, by addition,

A+K : K : : a+c : c ; and for the same reason,
B+L : L : : b+d : d, or, by inversion,
L : B+ L ; id: b+^. And, since it has been sh^wn,
that Kih: I e; d ; therefore, ex aequo,

K, A, B, L,

c, a, by dm

A4-K, K, L, B+L,
a + c» A ^9 b-^rd.

A-f-K : B+h : : or +c : h+d ; but by hypothesis,
A+ C : B+D : : a+c : 6+df, therefore
A+K r A+C :: B+L : B+D.

S23 NOTES. SPPPL. BOOK L

Now, first, let K and C be supposed equal, then it is evident that L
and D are also equal ; and therefore, since by construction a : c z t
A : K, we have also a : c : : A : C ; and, for the same reason, i:d::
B : D, and these analogies from the first of the two conditions, of
which one is affirmed above to be always essentia) to the truth of To-
relli's proposition.

Next, if K be greater than C, then, since
A+K : A+C : : B+L : B+D, by division,
A+K : K — C : : B+L : L — D. But, as was shewn,
K : L : : C : D, by conversion and alternation,
K — C : K : : L— D : L, therefore, ex aequo,
A+K : K : : B+L : L, and lastly, by division,
A : K : : B : L, or A : B : : K : L, that is,
A : B : : C : D.
Wherefore, in this case the ratio of A to B is equal to that of C to
D, and consequently, the ratio of a to 6 equal to that of c to d. The
same may be shdwn, if K is less than C ; therefore in every case
therie are conditions necessary to the truth of Torelli's proposition,
which he does not take into account, and which, as is easily shewn,
do not belong to the magnitudes to which he applies it.

In consequence of this, the conclusion which he meant to establish
respecting the circle, falls entirely to the. ground, and with it the
general inference aimed against the modern analysis.

It will not, [ hope, be imagined, that I have taken notice of these
circumstances with any design to lessen the reputation of the learned
Italian, who has in so many respects deserved well of the mathemati*
^al sciences, or to detract from the value of a posthumous work, which
by its elegance and correctness, does so much honour to the English
editors. But I would warn the student against that narrow spirit whiek
seeks to insinuate itself eveh into the abstractions of geometry, and
' would persuade us, that elegance, nay, truth itself, are possessed ex-
clusively by the ancient methods of demonstration. The high tone
in which Torelli censures the modern mathematics is imposing, as k
is assumed by on^ who had studied the writings of Archimedes with
uncommon diligence. His errors are onthat account the more dan*
gerous, and require to be the niore carefully pointed out.

PROP. IX. -

This enunciation is the same with that of the third of the Dimensio
Cireuli of Archimedes ; but the demonstration is different, though it
proceeds like that of the Greek Geometer, by the continual bisection
of the 6th part of the circumference.

The limits of the circumference are thus assigned ; and the method
of bringing it about, notwithstanding many quantities are neglected in
the arithmetical operations, that the errors shall iq one case be all on
the side of defect, and in another all on the side of excess (in which I
have followed Archimedes,) deserves particularly to be observed, as
aflbrding'a good intiodactioittothe general methodsof approximatioa*

NOTES. SDP?L. BOOK IL

823

BOOK II.

DEF. VIll. and PROP. XX.

Solid angles, which are defined here in the same manner as in
Euclid, are magnitudes of a very peculiar kind, and are particularly

^ to be remarked for not admitting of that accurate comparison, one
*with another, which is common in the other subjects of geometricai
investigation. It cannot, for example, be said of one solid angle, that
it is the half, or the double of another solid angle ; nor did any
geometer ever think.of proposing the problem of bisecting a given
«olid angle. In a word, no multiple or sub-multiple of such an angle
can be taken, and we have no way of expounding, even to the sim-
plest cases, the ratio which one of them bears to another.

In this respect, therefore, a solid angle differs from every other mag.
nitude that is the subject of mathematical reasoning, all of which have
this common properly, that multiples and sub-multiples of them may
be found. It is not our business here to inquire into the reason of
this anomaly, but it is plain, that on -account of it, our knowledge of
the nature and the properties of such angles can never be very far
extended, and that our reasonings concerning them must be chiefly
confined to the relations of the plane angles, by which they are cim*
tained. One pf the most remarkable of those relations is that which is
demonstrated in the 21st of this Book, and which is, that ali the plane
angles which contain any solid angle must together be less than four
right angles. This proposition is the 2i8t of the 11th of Euclid.

This proposition, however, is subject to a restriction in certain
cases, which, I believe, was first observed by M. le Sage of Geneva,
in a communication to the Academy of Sciences of Paris in 1756.
When the section of the pyramid formed by the planes that contain jthe
solid angle is a figure that has none of its angles exterior, such as a
triangle, a parallelogram, &c. the truth of the proposition just enunciat.
ed cannot be questioned. But, when the aforesaid section is a figure

iike that which is annexed, viz.
ABCD, having some angles such
as BDC, exterior, or, as they are
sometimes called, re-entering an-
gles, the proposition is not necessari-
ly true ; and It is plain, that in such

' cases the demonstration which we
have given, and which is the same
with Euclid's, will on longer apply.
Indeed, it were easy to, show, that- -n
on bases of this kind, by multiplying
the aamber of sides^ solid angles may be formed^ such that the plane

824 NOTlES. StJPPL. BOOK H.

angles which contain them shall exceed four right angles by any
quantity assigned. An illustration of this from the properties of the
sphere is perhaps the simplest of all others. Suppose that on the suiu
face of a hemisphere there is described a figure bounded by any num-
ber of arches of great circles making angles with one another, on op-
posite sides alternately, the plane angles at the centre of the sphere
that stand on these arche^ may evidently exceed four right angles,
and that too, by multiplying and extending the arches in any assigned
ratio. Now, these plane angles contain a solid angle at the centre
of the sphere, according to the definition of a solid angle.

We are to understand the proposition in the text, therefore, to be
true only of those solid ungles in which the inclination of the plane
angles are all the same way, or all directed toward the interior of the
figure. To distinguish this class of solid angles from that to which the
proposition does not apply, it is perhaps best to make use of this
criterion, that they are such, that when any two points whatsoever
ere taken in the planes that contain the solid angle, the straight line,
joining those points, falls wholly witliin the solid angle : or thus, they
are such, that a straight line cannot meet the planes which contain
them in more than two points. It is thus, too, that I would distinguish
a plane figure that has none of its angles exterior, by saying, that it
is a rectilineal figure, such that a straight line cannot meet the
boundary of it in more than two points.

We, therefore, distinguish solid angles into two species : one m
which the bounding planes can be intersected by a straight line only
in two points ; and another where the bounding planes may be inter-
sected by a straight line in more than two points : to the first of these
the proposition in the text applies, to the second it does not.

Whether Euclid meant entirely to exclude the consideration of
figures of the latter kind, in all that he has said of solids, and of solid
angles, it is not now easy to determine : it is certain, that his defini-
tions involve no such exclusion ; and as the introduction of any limita-
tion would considerably embarrass these definitions, and render them
difficult to be understood by a beginner, I have left it out, reserving to
this place a fuller explanation of the difficulty. I cannot conclude this
note without remarking, with the historian of the Academy, that it is
extremely singular,, that not one of all those who had read or ex-
plained Euclid before M. le Sage, appears to. have been sensible of
this mistake. (Memoires de V Acad, des Sciences, 1756, Hist. p. 77.)
A circumstance that renders this still more singular is, that another
mistake of Euclid on the same subject, and perhaps of all other,
geometers, escaped M. lo Sage also, and was first discovered by Dr.
Simson, as will presently appear. '^ .

PROP. IV,

Thid very elegant demonstration is from Legendte, and is much
easier than that of £uclid%

NOTES. SCPPL. BOOK IIL . 8SU

The demonstration giv^n here of the 6th is also greatly simpler
•than that of Euclid. It has even an advamage that does not belong
to Legendre's, ihat of requiring no particular construction or deter-
mination of any one of the lines, but reasoning from properties com-
mon to every pairt of them. The simplification, when it can be in-
irafduced, which, however, does not appear to be always possible, ii^,
-perhapSj the greatest improvement that can be made on an elemen-
tary demonstration.

PROP. XIX.

The problem contained in this proposition, of drawing a straight
4iae perpendicular to two straight lines not in the same plane, is cer-
tainly to be accounted elementary, although not given in any boo)^
of elementary geometry that I know of before that of Legendre.
The solution given here is more simple than his, or than any other
that I have yet met with : it also leads more easily, if it be required|
^to a trigonometrical computation.

BOOK III.

DEF. II. and PROP. I.

Tbbse relate to similar and^qual solids, a subject on which mis-
'tokes have prevailed not unlike to that which has just been mentioned*
The equAlity of solids, it is natural to expect, must be proved like thjO
-equality of plane figures, by showing that they may be made to coin-
cide, or to occupy the «ame space. But, though it be true that all
solids which can be shewn to coincide are equal and similar, yet it
d^ies not hold conversely, that all solids which are equal and similar
can be made to coincide. Though this assertion may appear some-
what paradoxical, yet the proof of it is extremely simple.

Let ABC be an isosceles triangle, of which the equal sides are AB
«nd AC ; from A draw A E perpendicular to the base BC, and BC will
be bisected in E. From E draw ED per- ^X>

pendicular to the flane ABC, and from D,
any point in it, draw DA, DB, DC to the
three angles of the triangle ABC. The
pyramid DABC is divided into two pyra-
mids DABE^DaCE, which, though their
equaKty ^will not be disputed, cannot be
flo applied to one another as to coincide.
For, though the triangles ABE, ACE are

equal, BE being equal to CE, EA common

t6 both, and the angles AEB, AEC equal, 5 £ C

^beeowe thdy- 9XQ right aingLes^ yeLit tb/ese

826 NOTES. SUPPL. BOOK HI.

two triangles be applied to one another, so as to coincide, thjB solid
DACE will nevertheless, as is evident, fall without the solid DA BR,
for the two solids' will be on the opposhe sides of the plane ABE.
In the same way, though all the planes of the pyramid DA BE may
easily be shewn to be equal to those of the pyramid DACE, each to
each ; yet will the pyramids themselves never coincide, though the
equal planes be applied to one another, because they are on the op-
posite sides of those planes.

It may be said, then, onwhat ground do we conclude the pyramids
to be equal ? The answer is, because their construction is entirely
the same, and the conditions that determine the magnitude of the one
identical with those that determine the magnitude of the other. For
the magnitude of the pyramid DABE is determined by the magni-
tude of the triangle ABE, the length of the line £D, and the position
of ED, in respect of the plane A BE ; three •ircumsl^nces that are
precisely the same in the two pyramids, so that there is nothing that
can determine one of them to be greater than another.

This reasoning appears perfectly conclusive and satisfactory ;< and
it seems also very certlin, that there is no other principle equally
simple* on which the relation of the solids DABE, DACE to one an-
other can be determined. Neither is this a case tHat occurs rarely ;
it is one, that, in the comparison of magnitudes having three dimen-
sions, presents itself continually ; for, though two plane figures that
are equal and similar can always be made to coincide, yet, with re-
gard to solids that are equal and similar, if they have not a certain
similarity in their position, there will be found just as many cases in'
which they cannot, as in which thej^ can coincide. Even figures de-
scribed on surfaces, if they are nut plane surfacc^s, may be equal and
similar without the possibility of coinciding. Thus, in the figure de-
scribed on the surface of a sphere, called a spherical triangle, if we
-suppose it to be isosceles, and a perpendicular to be drawn from the
vertex on the base, it wilt not be doubted, that it is thus divided into
two right angled spherical triangles equal and similar to one another,
and which, nevertheless, cannot be so laid on one another as to agree.
The same holds in innumerable other instances, and therefore it is
evident, that a principle, more general and fundamental than that of the
equality of coinciding figures, ought to be introduced into Geometry,
What this princiille is has also appeared very clearly in the course of
these remarks ; and it is indeed no other than the principle so cele-
brated in the philosophy of Leibnitz, under the name of the suffi-
ciBN*^ KBASON. For it was shewn, that the pyramids DABE and
DACE are concluded to be equal, because each of them is determined
to be of a certain magnitude, rather than of any other, by conditions
that are the same in both, so that there is no reason for the one being
greater than the other. This Axiom may be rendered general by
^saying. That things of which the magnitude is determined by condi-
tions that are exactly the same, are equal to one another ; or, it might
be expressed thus ; Two magnitudes A and B are equal, wheo there

NOTES. SUPPL. BOOK III. asW

is no reason that A should exceed B, rather than that B should exceed
A. Either of these will servo as the fundamental principle for com-
paring geometrical magnitudes of every kind ; they will apply in those
cases where the coincidence of magnitudes with one another has no^
place ; and they will apply with great readiness to the cases in which
a coincidence may take pUce, such as in the 4th, the 8th, or the 26th
pf the First Book of the Elements. '

The only oojection to this Axiom is, that it is somewhat of a meta-
physical kind, and belongs to the doctrine of the sufficient reason^ which
is looked on with a suspicious eye by somq philosophers. But this is
no solid objection ; for such reasoning may be applied with the great-
est safety to those objects with the nature of which we are perfectly
acqjuainted, and of which we have complete definitions, as in pure
mathematics. In physical questions, the same principle cannot be
applied with equal safety, because in such cases we have seldom a
complete definition of the thing we reason about,- or one that includes
all its properties. Thus, when Archimedes proved the spherical
figure of the earth, by reasoning on a principle of this sort, be was
led to a false conclusion, because he knew nothing of the rotation of
the earth on its a^is, which places the particles of that body, though
at equal distances from the centre, in circumstances very different
from one another* But, concerning those things that are the creatures
of the mind altogether, like the objects of mathematical investigation,
there can be no danger of being misled by the principle of the suf-
ficient reason, which at the same tima furnishes us with the only sin-
gle Axiom, by help of which we can compare together geometrical
quantities, whether they be of one, of t\^o, or of three dimensions.

Legend re in his Elements has made the same remark that has been
just stated, that thero are solids and other Geometrical Magnitudes,
which, though similar and equal, cannot be brought to coincide with
one another, and he has distinguished* them by the name of Symme"
triced Magnitudes. He has also given a very stUisfactory and iuge*
nioUs demonstration of the equality of certain solids of that sort,
though not so concise as the nature of a simple and elementary truth
would seem to require, and consequently not such as to render the
axiom proposed above altogether unnecessary.
• But a circumstance for which I cannot very well account is, that
Legendre, and after him Lacroix, ascribe to Simson the first mention
of such solids as we are here considering. Now I must be permitted
to say, that no remark to this' purpose is to be found in any of the wri-
tings of Simson, which have come to my knowledge. He has indeed
made an observation concerning the Geometry of Solids, which was
both new and important, viz. that solids may have the condition which
EucHd thought sufficient to determine their quality, and may never-
theless be unequal ; whereas the observation made here is, that so«
lids may be equal and similar, and may yet want the condition of be.
ing able to coincide with one another. These propositions are widely
different ; and how so accurate a writer as Legendre should have mi»-

N I

NOTES. SUPPL. BOOK lit .

taken the one for the other, is not easy to be explained. It must be
observed, that he does not seem in the least aware of the observation
which Simson has really made. - Perhaps having himself made the re*
mark we now speak of, and on looking slightly into Simson, having
found a limitation of the usual description of equal solids, he had with-
out much inquiry, set it down as the same, with his own notion ; and
so. With a great deal of candour, and some precipitation, he has as-
oribed to Simson a discovery which really belonged to Himself. This
at least seems to be the most probable solution of the difficulty.

I have enteried into a fuller discussion of Legendre's mistake thaa
I should otherwise have done, from having said, in the first edition of
these elements, in 1795, that I believed the non-coincidence of simi->
lar and equal solids in certain circumstances, was then mede for the
first time. This it is evident would have been a pretension as ridi-
culous as ill-founded, if the same observation had been made in a book
like Simson's, which in this country was in every body's hands, and
which I had myself professedly studied with attention. As I have
not seen any edition of Legendre's Elements earlier than that pub-
lished in 1802, 1 am ignorant whether he or I was the first in. making
&e remark here referred to. That circumstance is, however, imma-
lertal ; for I am not interested about the originality of the remaifcj
though very much interested to show that I bad no intention of ap-
propriating to myself a discovery made by another*

Another observation on the subject of those solids, wbich, with Le-
gendre, we shall call Symmetrical, has occurred to me, which J did
not at first think of, viz. that Euclid himself certainly had these solids
in view when he formed his definition (as he very improperly calls it)
of equal and similar solids. He says that those solids are equal and
similar J which are contained under the same number of equal and simi*
lar planes. But this is not true, as Dr. Simson has shewn in a passage
just about to be quoted, because two solids may easily be assignedy
bounded by the same number of equal and similar planes, which are
obviously unequal, the one being contained within the other. Simson
observes, that Euclid needed only to have added, that the equal and
similar planes must be similarly situated, to have made his description
exact. Now, it is true, that this addition would have made it exact
in one respect, but'would have rendered it imperfect in another ;. for
though all the solids having the conditions here enumerated, fire equal
and similar, many others are equal and similar which have not thoae
conditions, that is, though bounded by the same equal number of similar
planes, those planes are not similarly situated. The symmetrical Bolids
have not their equal and similar planessimilarly situated, but inan order
and position directly contrary. Euclid, it is probable, was aware of
thiSj and by seeking to render the description ofequal and similar so-
lids so general, as to comprehend solids of both kinds, bas stript it of
an essential condition, so that solids obviously unequal kte included
in it, and has also been led into a very illogical proceedbg, that of
defining the equality of solids, instead of proving it^a&if lie hftdiuBen

NOTES, SUPPL. BOOK III. 329

nt liberty to fix a new idea to the word equal every time thaKihe ap.
iplied it to a neW kind of magnitude. The nature of the difficulty he
had to contend with, will perhaps be the more readily admitted as an
apology for this error, when it is considered that Simson, who had
studied the matter so carefully, as to set Euclid right in one particu-
lar, was himself wrong in another, and has treated of equal and
similar solids, so as to exclude the symmetrical altogether, to which
indeed he seems not to have at all adverted.

I must, therefore, again repeat, that I do not think that this matter
can be treated in a way quite simple and elementary, and at the same
time general, without introducing the principle of the sufficient reasim
as stated above. It may then be demonstrated, that similar and equal
solids are those contained by the same number of equal and similar
planes, either with similar or contrary situations. If the word contra"
ry is properly understood, this description seems to be quite general.
Simson's remark, that solids may be unequal, though contained by
the same number of. equal and similar planes, extends also to solid
angles which may be unequal, though contained by the same number
of equal plane angles. These remarks he published in the first edi.
tion of his Euclid iii 1756, the very same year that M. le Sage com-
tBunicated to the Academy of Sciences the observation on the sub-
ject of solid angles, mentioned in a former note ; and it is singular,
that these two Geometers, without any communication with one ano-
ther, should almost at the same time have made two discoveries very
nearly connected, yet neither of them comprehending the whole
truth, so that each is imperfect without the other.

Dr. Simson has shewn the truth of his remark; by the following
reasoning.

^<Let tliere be any plane rectilineal figure, as the triangle ABC, and
from a point D within it, draw the straight line DE at right angles to
the plane ABC; in DE take DE, DF equal to one another, upon the
opposite sides of the plane, and let G be any point in EF ; join DA,
DB, DC ; EA, EB, EC ; FA, FB, FC ; GA, GB, GC : Because the
straight line EDF is at right angles to the plane ABC, it makes right
angles with DA, DB, DC, which it m^ets in that plane ; and in the
triangles EDB, FDB, ED and DB are equal to FD and DB, each to
each, and they contain right angles ; therefore the base EB is equal
to the basis FB ; in the same manner EA is equal to FA, and EC to
FC : and in the triangles EBA, FBA, EB, BA are equal to FB, BA,
and the base EA is equal to the base FA ; wherefore the angle EBA
is equal to the angle FBA, and the triangle EBA equal to the triangle
FBA, and the other angles equal to the other angles ; therefore these
triangles are similar: In the same manner the triangle EBC is similkr
to the triangle FBC, and the triangle EAC to FAC ; therefore there are
two soUd figures, each of which is contained by six triangles, one of
them by three triangles, the common vertex of which is the point G,
and their bases the straight lines AB, BC, CA, and by three other tri-
angles the common vertex of which is the point E, and their bases the

930

NOTES. SUPPL. BOOK IH.

same lines A6, BC, CA. The other solid is contained by the same
three triangles, the common vertex of which is G, and their bases
A6, BC, CA ; and by three other triangles, of which the common
vertex is the point F, and their bases the same straight lines AB, BC,
CA : Now, the three triangles GAB, GBC, GCA are common to both
solids, and the three others EAB, EBC, EC A, of the first solid have
been shown to be equal and similar to the three others,^FAB, FBC,
FCA of the other solid, each to each ; therefore, these two solids are
contained by the same number of equal and similar planes : But that
they are not equal is manifest, because the first of them is contained
in the other : Therefore it is not universally true, that solids are equal
which are contained by the same number of equal and similar planes."

'^ Cor. From this it appears, that two unequal solid angles may
be contained by the same number of equal plane angles."

" For the solid angle at B, which is contained by the four plane
angles EB A, EBC, GBA, GBC is not equal to the solid angle at the
same point B, which is contained by the four plane angles FBA, FBC,
GBA, GBC ; for the last contains the other. And each of them is
contained by four plane angles, which are equal to one another, each
to each, or are the self-same, as has been proved : And indeed, th«re
may be innumerable solid angles all unequal to one another, which
are each of them contained by plane angles that are equal to oneano-
other, each to each. It is likewise manifest, that the before-mention-
ed solids are not similar, since their solid angles are not all equal.''

NOTES. PL. TRIGONOMETRY. 331

PLANE TRIGONOMETRY,

DEFINITIONS, &c.

Trioonohetrt is defined in the text to be the application of
Number to express the relations of the sides and angles of tri-
angles. It depends, therefore, on the 47th of the first of Euclid, and
on the 7th of the first of the Supplement, the two propositions which
do most immediately connect together the sciences of Arithmetic
and Geometry.

The sine of an angle is defined above in the usual way, viz. the
perpendicular drawn from one extremity of the arch, which measures
the angle on the radius passing through the other ; but in strictness
the sine is not the perpendicular itself^ but the ratio of that perpen-
dicular to the radius, for it is this ratio which remains constant, while
the angle continues the same, though the radius vary. It migh^ be
convenient, therefore, to define the sine to be the quotient which
arises from dividing the perpendicular just deseribed by the radius
of the circle.

So also, if one of the sides of a right angled triangle about the right
ongle be divided by the other, the quotient is the tangent of the angle
opposite to the first-mentioned side, &c. But though this is certainly
the rigorous way of conceiving the sines, tangents, 6sc. of angles,
which are in reality not magnitudes, but the ratios of magnitudes ;
yet as this idea is a little more abstract than the common one, and
would also involve some change in the language of trigonometry, at
the same time that it would in the end lead to nothing that is not at->
tained by making the* radius equal to unity, I have adhered to the
common method, though I have thought it right to point out that
which should in strictness be pursued.

A proposition is \e& out in the Plane Trigonometry, which the
astronomers make use of in order, when two sides of a triangle, and
the angle contained by them, are given, to find the angles at the
base, without making use of the sum or difierence of the sides, which,
in some cases, when only the Logarithms of the sides are given, can-
not be conveniently found.

8^ NOTES. PL. TRIGONOMETRY.

THEOREM.

If^ as the greater of any two sides of a triangle to the less, so the raiia»
to the tangent of a certain angle; then mil the radius be to the tan-
gent of the difference between that angle and half a right angle, as the
tangent of half the sum <fthe angles, atthe hereof the triangletoihe
tangent of half their difference.

Let ABC be a triangle, the sides of which are BC and C A, and the
base AB, and let BC be greater than ^ I

CA. Let DC be drawn at right '^

angles to BC, and eqnal to AC ;
join- BD, and because (Prop. 1.)
in the right angled triangle BCD,
BC : CD : : R : tan CBD, CBD is
the angle of which the tangent is D
to the radius as CD to BC, that is,
as C A to BC, or as the least of the
two sides of the triangle to the
greatest.

But BC+CD : BC-CD : : tan i (CDB+CBD) :
tan i (CDB-CBD; (Prop. 5.) ;
and also, BC+CA : BC-CA : : tan ^ (CAB+CBA) :
tan i (CAB— CBA). Therefore, since CD=CA,
tan i (CDB+CBD) : tan i (CDB-CBD) ::
tan i (CAB+CBA) : tan i (CAB— CBA). But because the
angles CDB+CBD =9(F, tan i (CDB+CBD) :
tan i (CDB-CBD) : : R : tan (45o-CBD), (2 Cor. Prop. 3.) ;
therefore, R : tan (45°— CBD) : : tan i (CAB+CBA) :
tan i (CAB — CBA) ; and CBD was already shewn to be such an an-
gle that BC : CA : : R : tan CBD. Therefore, &c. Q, E. D.

Cor. If BC, CA, and the angle C are given to find the angles A
and B ; find an angle £ such, that BC : CA : : R : tan E ; then R :
tan (45°— B) : : tan i (A+B) : tan i (A— B). Thus i (A— B) is
found, and i (A+B) being given, A and B are each of them known,
Lem. 2.

In reading the elements of Plane Trigonometry, it may be of use
to observe, that the first five propositions contain all the rules abso-
lutely necessary for solving the different cases of plane triangles.
The learner, when he studies Trigonometry for the first time, may
satisfy himself with these propositions, but should by no means neg-
lect the others in a subsequent perusal.

PROP. VIL and VIII.

I have changed the demonstration which I gave of these proposi-
tions in the first edition, for two others considerably simpler and more
concise, given me by Mr. Jardine, teacher of the Mathematics in
Edinburgh, formerly one of my pupils, to whose ingenuity and skill I
am very glad to bear this public testimony.

NOTES. SPH. TRI60N0METRT.

SPHERICAL

TRIGONOMETRY,

iPROP. V.

Ths angle! at the base of an iaosceles spherical triangle are symf
metrical magnitudes, not admitting of being laid on one another, nor
of coinciding, notwithstanding their equality. It might be consider-
ed as a sufficient proof that they are equal, toobserve that they are
each deternyned to be of a certain magnitude rather than any other,
by conditions which are precisely the same, so that there is no rea-
son why one of them should be greater than anpther. ,For the sake
of those to whom this reasoning may not prove satisfisustory, the de-
monstration in the text is given, which is strictly geometrical.

For the demonstrations of the two propositions tbit are given in the
end of the Appendix to the Spherical Trigonometry, see Elementa
Sphffiricorum, Theor. 66. apud Wolfii Opera Math. tom. ili. ; Trigo-
nometric par Cagnoli, § 463 ; Trigonometrie Spherique par Mauduit,
5 165.

FINIS.

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